CBSE 2024 · Region 4 · Set 1 · Q23 · 3 marks
Calculate emf of the following cell at $\displaystyle 25$ °C : $\displaystyle \mathrm{Sn} / \mathrm{Sn}^{2+}(0.001 \mathrm{M}) \| \mathrm{H}^{+}(0.01 \mathrm{M})\left|\mathrm{H}_{2(\mathrm{~g})}(1 \mathrm{bar})\right| \mathrm{Pt}_{(\mathrm{s})}$ Given : $\displaystyle \mathrm{E}^{\circ}\left(\mathrm{Sn}^{2+} / \mathrm{Sn}\right)=-0.14 \mathrm{~V}, \mathrm{E}^{\circ} \mathrm{H}^{+} / \mathrm{H}_{2}=0.00 \mathrm{~V}(\log 10=1)$ $\displaystyle 3$
Marking-scheme solution
\(\displaystyle \mathrm{E}_{\text {cell }}=\mathrm{E}^{\mathrm{o}}{ }_{\text {cell }}-\frac{0.059}{2} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}}\)
\[\mathrm{E}_{\text {cell }}^{\mathrm{o}}=0-(-0 \cdot 14 \mathrm{~V})=0 \cdot 14 \mathrm{~V}
\]
\[\begin{aligned}
& \mathrm{E}_{\text {cell }}=0 \cdot 14-\frac{0 \cdot 059}{2} \log \frac{(0 \cdot 001)}{(0 \cdot 01)^{2}} \\
& =0 \cdot 14-\frac{0 \cdot 059}{2} \log 10 \\
& =0 \cdot 14-0 \cdot 0295=0 \cdot 1105 \mathrm{~V} \text { or } 0.11 \mathrm{~V}
\end{aligned}
\]
ElectrochemistryNernst EquationApplyshort_answermedium
More from Electrochemistry
- Kohlrausch gave the following relation for strong electrolytes: Λ m=Λ m^°-A √C Which of the following…2025 · asked 4×
- (i) What should be the signs (positive/negative) for E^° Cell and Δ G^° for a spontaneous redox reaction…2023 · asked 3×
- Batteries and fuel cells are very useful forms of galvanic cell. Any battery or cell that we use as a source…2024 · asked 3×
- (i) Calculate emf of the following cell at 25°C: Zn( s) Zn^2+(0 · 001 M) Cd^2+(0 · 1 M) Cd( s) Given: E Zn^2+…2024 · asked 3×
- In a galvanic cell, chemical energy of a redox reaction is converted into electrical energy, whereas in an…2024 · asked 3×
- (i) For a galvanic cell, the following half reactions are given. Decide, which will remain as reduction…2025 · asked 3×
- Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: ( 3 + 2 = 5 ) Sn(s)…2025 · asked 3×
- Four half reactions I to IV are shown below: I. 2 Cl l^- → Cl 2+2 e^- II. 4 OH^- → O 2+2 H 2 O+2 e^- III.…2023 · asked 3×
CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.