Calculate the emf of the following cell at 25^ ^ C Zn(s) Zn^ ++ (0.1M) Cu^ ++ (0.05M) Cu(s) if E^ ^ cells 1.1V.

Correct answer: A. 1.091 V. The cell reaction is Zn(s) + Cu^ 2+ (aq) arrow Zn^ 2+ (aq) + Cu(s) Number of electrons transferred, n = 2 Using the Nernst equation: E_ cell = E^ ^ _ cell - (0.0591)/(n) [Zn^ 2+ ] [Cu^ 2+ ] Substituting the given values: E_ cell = 1.1 - (0.0591)/(2) (0.1)/(0.05) E_ cell = 1.1 - 0.02955 2 E_ cell = 1.1 - 0.02955 × 0.3010 E_ cell = 1.1 - 0.0089 E_ cell = 1.0911 V Answer: 1.091 V.

MHT CET 202621 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

MHT CET 2026 · 21 April · Morning Shift · Chemistry · Electrochemistry

Calculate the emf of the following cell at 25^ ^ C Zn(s) Zn^ ++...

Calculate the emf of the following cell at

25^{}^{\circ}\text{C}

\text{Zn(s)} \mid \text{Zn}^{++}(0.1\text{M}) \parallel \text{Cu}^{++}(0.05\text{M}) \mid \text{Cu(s)}

E^{}^{\circ}

cells 1.1V.

Options

A1.091 V.
B1.2 V.
C1.52 V.
D1.41 V

Correct answer

A. 1.091 V.

Step-by-step solution

The cell reaction is

\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}

Number of electrons transferred,

n = 2

Using the Nernst equation:

E_{\text{cell}} = E^{{}^{\circ}}_{\text{cell}} - \dfrac{0.0591}{n} \log \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Substituting the given values:

E_{\text{cell}} = 1.1 - \dfrac{0.0591}{2} \log \dfrac{0.1}{0.05}

E_{\text{cell}} = 1.1 - 0.02955 \log 2

E_{\text{cell}} = 1.1 - 0.02955 \times 0.3010

E_{\text{cell}} = 1.1 - 0.0089

E_{\text{cell}} = 1.0911 \text{ V}

Answer: 1.091 V.

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