Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Electro Chemistry

Question:

A cell is prepared by dipping a \(Cu\) rod in \(\text{1 M }CuSO_4\) solution and \(Sn\) rod in \(\text{1 M }SnCl_2\) solution. The standard electrode potential of \(Cu\) is \(+0.34 V\) and Sn is \(-0.14 V\). Emf of cell will be:

Options:

\(0.48 V\)

\(0.20 V\)

\(0.34 V\)

\(0.14 V\)

Correct Answer:

\(0.48 V\)

Explanation:

The correct answer is option 1. \(0.48 V\)

To calculate the electromotive force \((emf)\) of an electrochemical cell, you need to understand the components of the cell and how they contribute to the overall voltage. The emf of a cell is a measure of the cell's ability to push electrons through an external circuit.

In this scenario, you have a cell consisting of a copper \((Cu)\) rod and a tin \((Sn)\) rod, each immersed in its respective solution: a 1 M \(CuSO_4\) solution for the \(Cu\) rod and a 1 M \(SnCl_2\) solution for the Sn rod. You are given the standard electrode potentials for \(Cu\) and \(Sn\) as \(+0.34\, \ V\) and \(-0.14\, \  V\), respectively.

Let us break down the cell components and the reactions that occur at each electrode:

1. Cathode (Cu Rod):

\(Cu\) metal undergoes oxidation at the electrode:

\(Cu \rightarrow Cu^{2+} + 2e^-\)

\(Cu\) metal loses two electrons and forms \(Cu^{2+}\) ions.

2. Anode (Sn Rod):

\(Sn^{2+}\) ions in the \(SnCl_2\) solution undergo reduction at the electrode:

\(Sn^{2+} + 2e^- \rightarrow Sn\)

\(Sn^{2+}\) ions gain two electrons and form \(Sn\) metal.

Now, let's calculate the emf of the cell using the formula:

\(Emf_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\)

where

\(E_{\text{cathode}}\) is the standard electrode potential of the cathode (Cu in this case), which is +0.34 V

\(E_{\text{anode}}\) is the standard electrode potential of the anode (Sn in this case), which is -0.14 V.

Substitute these values into the formula:

\(Emf_{\text{cell}} = (+0.34 \, \text{V}) - (-0.14 \, \text{V})\)

Now, simplify:

\(Emf_{\text{cell}} = 0.34 \, \text{V} + 0.14 \, \text{V} = 0.48 \, \text{V}\)

So, the emf of the cell is 0.48 V. This positive value indicates that the cell can supply electrical energy to an external circuit and is a measure of the cell's ability to do work.