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Redox Reactions and Electrochemistry: Chemistry | JEE Main P t ( s ) ∣ H 2 ( g ) ( 1 b a r ) ∣ ∣ H + ( a q ) ( 1 M ) ∥ M 3 + ( a q ) , M + ( a q ) ∣ P t ( s ) \mathrm{Pt(s)}|\mathrm{H}_2(\mathrm{g})(1\mathrm{bar})||\mathrm{H}^+(\mathrm{aq})(1\mathrm{M})\parallel\mathrm{M}^{3+}(\mathrm{aq}), \mathrm{M}^+(\mathrm{aq})|\mathrm{Pt(s)} Pt ( s ) ∣ H 2 ( g ) ( 1 bar ) ∣∣ H + ( aq ) ( 1 M ) ∥ M 3 + ( aq ) , M + ( aq ) ∣ Pt ( s )
The
E c e l l \mathrm{E}_{\mathrm{cell}} E cell for the given cell is
0.1115 V 0.1115\text{ V} 0.1115 V at
298 K 298\text{ K} 298 K when
[ M + ( a q ) ] [ M 3 + ( a q ) ] = 10 a \frac{[\mathrm{M}^+(\mathrm{aq})]}{[\mathrm{M}^{3+}(\mathrm{aq})]} = 10^a [ M 3 + ( aq )] [ M + ( aq )] = 1 0 a
The value of
a a a is
Given :
E 0 M 3 + / M + = 0.2 V \mathrm{E}^0\mathrm{M}^{3+}/\mathrm{M}^+ = 0.2\text{ V} E 0 M 3 + / M + = 0.2 V 2.303 R T F = 0.059 V \frac{2.303\mathrm{RT}}{\mathrm{F}} = 0.059\text{ V} F 2.303 RT = 0.059 V Hint 1 of 3
What is the overall cell reaction and the number of electrons transferred (n n n
1 2 H 2 ( g ) + M 3 + ( a q ) → H + ( a q ) + M + ( a q ) \frac{1}{2}\mathrm{H}_2(\mathrm{g}) + \mathrm{M}^{3+}(\mathrm{aq}) \rightarrow \mathrm{H}^+(\mathrm{aq}) + \mathrm{M}^+(\mathrm{aq}) 2 1 H 2 ( g ) + M 3 + ( aq ) → H + ( aq ) + M + ( aq ) n = 1 n = 1 n = 1 H 2 ( g ) + M 3 + ( a q ) → 2 H + ( a q ) + M + ( a q ) \mathrm{H}_2(\mathrm{g}) + \mathrm{M}^{3+}(\mathrm{aq}) \rightarrow 2\mathrm{H}^+(\mathrm{aq}) + \mathrm{M}^+(\mathrm{aq}) H 2 ( g ) + M 3 + ( aq ) → 2 H + ( aq ) + M + ( aq ) n = 2 n = 2 n = 2 No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
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Choose one answer I’d rather see the answer directly Step-by-step solution View Correct answer
The value of a a a Option analysis
Why each option works or fails
Step Working
01 given Cell: P t ( s ) ∣ H 2 ( g ) ( 1 bar ) ∣ H + ( a q ) ( 1 M ) ∥ M 3 + ( a q ) , M + ( a q ) ∣ P t ( s ) \mathrm{Pt(s)}|\mathrm{H}_2(\mathrm{g})(1\text{ bar})|\mathrm{H}^+(\mathrm{aq})(1\text{ M}) \parallel \mathrm{M}^{3+}(\mathrm{aq}), \mathrm{M}^+(\mathrm{aq})|\mathrm{Pt(s)} Pt ( s ) ∣ H 2 ( g ) ( 1 bar ) ∣ H + ( aq ) ( 1 M ) ∥ M 3 + ( aq ) , M + ( aq ) ∣ Pt ( s ) E cell = 0.1115 V E_{\text{cell}} = 0.1115\text{ V} E cell = 0.1115 V 298 K 298\text{ K} 298 K [ M + ] [ M 3 + ] = 10 a \frac{[\mathrm{M}^+]}{[\mathrm{M}^{3+}]} = 10^a [ M 3 + ] [ M + ] = 1 0 a E M 3 + / M + ∘ = 0.2 V E^\circ_{\mathrm{M}^{3+}/\mathrm{M}^+} = 0.2\text{ V} E M 3 + / M + ∘ = 0.2 V 2.303 R T F = 0.059 V \frac{2.303RT}{F} = 0.059\text{ V} F 2.303 R T = 0.059 V
02 find Find the value of the integer/exponent a a a
03 strategise 1. Write the half-cell reactions and overall cell reaction:
Anode (oxidation): H 2 ( g ) ⟶ 2 H + ( aq ) + 2 e − \mathrm{H}_2(\text{g}) \longrightarrow 2\mathrm{H}^+(\text{aq}) + 2e^- H 2 ( g ) ⟶ 2 H + ( aq ) + 2 e − E ∘ = 0.0 V E^\circ = 0.0\text{ V} E ∘ = 0.0 V M 3 + ( aq ) + 2 e − ⟶ M + ( aq ) \mathrm{M}^{3+}(\text{aq}) + 2e^- \longrightarrow \mathrm{M}^+(\text{aq}) M 3 + ( aq ) + 2 e − ⟶ M + ( aq ) E ∘ = 0.2 V E^\circ = 0.2\text{ V} E ∘ = 0.2 V n = 2 n = 2 n = 2 H 2 ( g ) + M 3 + ( aq ) ⟶ 2 H + ( aq ) + M + ( aq ) \mathrm{H}_2(\text{g}) + \mathrm{M}^{3+}(\text{aq}) \longrightarrow 2\mathrm{H}^+(\text{aq}) + \mathrm{M}^+(\text{aq}) H 2 ( g ) + M 3 + ( aq ) ⟶ 2 H + ( aq ) + M + ( aq ) E cell ∘ = E cathode ∘ − E anode ∘ = 0.2 − 0 = 0.2 V E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.2 - 0 = 0.2\text{ V} E cell ∘ = E cathode ∘ − E anode ∘ = 0.2 − 0 = 0.2 V E cell = E cell ∘ − 0.059 n log Q E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q E cell = E cell ∘ − n 0.059 log Q Q = [ M + ] [ H + ] 2 [ M 3 + ] ⋅ P H 2 = [ M + ] [ M 3 + ] = 10 a Q = \frac{[\mathrm{M}^+][\mathrm{H}^+]^2}{[\mathrm{M}^{3+}]\cdot P_{\mathrm{H}_2}} = \frac{[\mathrm{M}^+]}{[\mathrm{M}^{3+}]} = 10^a Q = [ M 3 + ] ⋅ P H 2 [ M + ] [ H + ] 2 = [ M 3 + ] [ M + ] = 1 0 a a a a
04 execute 0.1115 = 0.2 − 0.059 2 log ( 10 a ) 0.1115 = 0.2 - \frac{0.059}{2}\log(10^a) 0.1115 = 0.2 − 2 0.059 log ( 1 0 a ) 0.059 2 ⋅ a = 0.2 − 0.1115 = 0.0885 \frac{0.059}{2} \cdot a = 0.2 - 0.1115 = 0.0885 2 0.059 ⋅ a = 0.2 − 0.1115 = 0.0885 a = 2 × 0.0885 0.059 = 0.1770 0.059 = 3 a = \frac{2 \times 0.0885}{0.059} = \frac{0.1770}{0.059} = 3 a = 0.059 2 × 0.0885 = 0.059 0.1770 = 3
✓ verify Check: 0.059 2 × 3 = 0.0885 \frac{0.059}{2} \times 3 = 0.0885 2 0.059 × 3 = 0.0885 0.2 − 0.0885 = 0.1115 V 0.2 - 0.0885 = 0.1115\text{ V} 0.2 − 0.0885 = 0.1115 V E cell E_{\text{cell}} E cell
Hints that build this answer step by step What is the overall cell reaction and the number of electrons transferred (n n n
H 2 ( g ) + M 3 + ( a q ) → 2 H + ( a q ) + M + ( a q ) \mathrm{H}_2(\mathrm{g}) + \mathrm{M}^{3+}(\mathrm{aq}) \rightarrow 2\mathrm{H}^+(\mathrm{aq}) + \mathrm{M}^+(\mathrm{aq}) H 2 ( g ) + M 3 + ( aq ) → 2 H + ( aq ) + M + ( aq ) n = 2 n = 2 n = 2 Given that the standard hydrogen electrode has E 0 = 0 V E^0 = 0\text{ V} E 0 = 0 V E c e l l 0 E^0_{\mathrm{cell}} E cell 0 Q Q Q
E c e l l 0 = 0.2 V E^0_{\mathrm{cell}} = 0.2\text{ V} E cell 0 = 0.2 V Q = [ H + ] 2 [ M + ] p H 2 [ M 3 + ] = [ M + ] [ M 3 + ] Q = \frac{[\mathrm{H}^+]^2[\mathrm{M}^+]}{p_{\mathrm{H}_2}[\mathrm{M}^{3+}]} = \frac{[\mathrm{M}^+]}{[\mathrm{M}^{3+}]} Q = p H 2 [ M 3 + ] [ H + ] 2 [ M + ] = [ M 3 + ] [ M + ] Substitute the given values into the Nernst equation E c e l l = E c e l l 0 − 0.059 n log Q E_{\mathrm{cell}} = E^0_{\mathrm{cell}} - \frac{0.059}{n}\log Q E cell = E cell 0 − n 0.059 log Q a a a a a a
a = 3 a = 3 a = 3 Your next move We think you should solve this next ✓ Source and academic review↓
Question type Numerical
Exam relevance JEE Main · Chemistry
Concepts assessed Chemistry
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why is n = 2 in the Nernst equation? Because the cathode reaction is M 3 + + 2 e − → M + \mathrm{M}^{3+} + 2e^- \to \mathrm{M}^+ M 3 + + 2 e − → M + H 2 → 2 H + + 2 e − \mathrm{H}_2 \to 2\mathrm{H}^+ + 2e^- H 2 → 2 H + + 2 e −
Answer The value of a a a
Why each option works or fails Step-by-step solution given: Cell: P t ( s ) ∣ H 2 ( g ) ( 1 bar ) ∣ H + ( a q ) ( 1 M ) ∥ M 3 + ( a q ) , M + ( a q ) ∣ P t ( s ) \mathrm{Pt(s)}|\mathrm{H}_2(\mathrm{g})(1\text{ bar})|\mathrm{H}^+(\mathrm{aq})(1\text{ M}) \parallel \mathrm{M}^{3+}(\mathrm{aq}), \mathrm{M}^+(\mathrm{aq})|\mathrm{Pt(s)} Pt ( s ) ∣ H 2 ( g ) ( 1 bar ) ∣ H + ( aq ) ( 1 M ) ∥ M 3 + ( aq ) , M + ( aq ) ∣ Pt ( s ) E cell = 0.1115 V E_{\text{cell}} = 0.1115\text{ V} E cell = 0.1115 V 298 K 298\text{ K} 298 K [ M + ] [ M 3 + ] = 10 a \frac{[\mathrm{M}^+]}{[\mathrm{M}^{3+}]} = 10^a [ M 3 + ] [ M + ] = 1 0 a E M 3 + / M + ∘ = 0.2 V E^\circ_{\mathrm{M}^{3+}/\mathrm{M}^+} = 0.2\text{ V} E M 3 + / M + ∘ = 0.2 V 2.303 R T F = 0.059 V \frac{2.303RT}{F} = 0.059\text{ V} F 2.303 R T = 0.059 V find: Find the value of the integer/exponent a a a strategise: 1. Write the half-cell reactions and overall cell reaction:
Anode (oxidation): H 2 ( g ) ⟶ 2 H + ( aq ) + 2 e − \mathrm{H}_2(\text{g}) \longrightarrow 2\mathrm{H}^+(\text{aq}) + 2e^- H 2 ( g ) ⟶ 2 H + ( aq ) + 2 e − E ∘ = 0.0 V E^\circ = 0.0\text{ V} E ∘ = 0.0 V M 3 + ( aq ) + 2 e − ⟶ M + ( aq ) \mathrm{M}^{3+}(\text{aq}) + 2e^- \longrightarrow \mathrm{M}^+(\text{aq}) M 3 + ( aq ) + 2 e − ⟶ M + ( aq ) E ∘ = 0.2 V E^\circ = 0.2\text{ V} E ∘ = 0.2 V n = 2 n = 2 n = 2 H 2 ( g ) + M 3 + ( aq ) ⟶ 2 H + ( aq ) + M + ( aq ) \mathrm{H}_2(\text{g}) + \mathrm{M}^{3+}(\text{aq}) \longrightarrow 2\mathrm{H}^+(\text{aq}) + \mathrm{M}^+(\text{aq}) H 2 ( g ) + M 3 + ( aq ) ⟶ 2 H + ( aq ) + M + ( aq ) E cell ∘ = E cathode ∘ − E anode ∘ = 0.2 − 0 = 0.2 V E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.2 - 0 = 0.2\text{ V} E cell ∘ = E cathode ∘ − E anode ∘ = 0.2 − 0 = 0.2 V E cell = E cell ∘ − 0.059 n log Q E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q E cell = E cell ∘ − n 0.059 log Q Q = [ M + ] [ H + ] 2 [ M 3 + ] ⋅ P H 2 = [ M + ] [ M 3 + ] = 10 a Q = \frac{[\mathrm{M}^+][\mathrm{H}^+]^2}{[\mathrm{M}^{3+}]\cdot P_{\mathrm{H}_2}} = \frac{[\mathrm{M}^+]}{[\mathrm{M}^{3+}]} = 10^a Q = [ M 3 + ] ⋅ P H 2 [ M + ] [ H + ] 2 = [ M 3 + ] [ M + ] = 1 0 a a a a execute: 0.1115 = 0.2 − 0.059 2 log ( 10 a ) 0.1115 = 0.2 - \frac{0.059}{2}\log(10^a) 0.1115 = 0.2 − 2 0.059 log ( 1 0 a ) 0.059 2 ⋅ a = 0.2 − 0.1115 = 0.0885 \frac{0.059}{2} \cdot a = 0.2 - 0.1115 = 0.0885 2 0.059 ⋅ a = 0.2 − 0.1115 = 0.0885 a = 2 × 0.0885 0.059 = 0.1770 0.059 = 3 a = \frac{2 \times 0.0885}{0.059} = \frac{0.1770}{0.059} = 3 a = 0.059 2 × 0.0885 = 0.059 0.1770 = 3 verify: Check: 0.059 2 × 3 = 0.0885 \frac{0.059}{2} \times 3 = 0.0885 2 0.059 × 3 = 0.0885 0.2 − 0.0885 = 0.1115 V 0.2 - 0.0885 = 0.1115\text{ V} 0.2 − 0.0885 = 0.1115 V E cell E_{\text{cell}} E cell