Question

In the electrochemical cell :
\(Zn|ZnSO_4(0.01M)||CuSO_4(1.0 M)|Cu,\)
the emf of this Daniel cell is \(E_1\). When the concentration of \(ZnSO_4\) is changed to 1.0 M and that of \(CuSO_4\) changed to 0.01 M, the emf changes to \(E_2\). From the following, which one is the relationship between \(E_1\) and \(E_2\)?
(Given, \(\frac {RT}{F}= 0.059\))

• A. \(E_1 = E_2\)
• B. \(E_1 < E_2\)
• C. \(E_1 > E_2\)
• D. \(E_2=0 ≠ E_1\)

Answer 1

The Correct Option is C

To find the relationship between the emf values \(E_1\) and \(E_2\) for the given electrochemical cell configurations, we need to apply the Nernst equation:

The Nernst equation for a cell is given by: 

\[E = E^\circ - \frac{RT}{nF} \ln Q\]

Where:

• \(E^\circ\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged in the electrochemical cell reaction.
• \(F\) is Faraday's constant.
• \(Q\) is the reaction quotient.

For the Daniel cell given:

\[\text{Zn}| \text{ZnSO}_4(a_1) || \text{CuSO}_4(a_2) | \text{Cu}\]

The net cell reaction is:

\[\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}\]

Thus, for both cases:

Calculation of \(E_1\):

In the initial condition: \(a_1 = 0.01 \, \text{M}\), \(a_2 = 1.0 \, \text{M}\)

\[E_1 = E^\circ - \frac{0.059}{2} \log \left( \frac{0.01}{1.0} \right)\]

 

\[E_1 = E^\circ + 0.059 \, \log(100)\]

 

\[E_1 = E^\circ + 0.118 \, \text{V}\]

Calculation of \(E_2\):

In the changed condition: \(a_1 = 1.0 \, \text{M}\), \(a_2 = 0.01 \, \text{M}\)

\[E_2 = E^\circ - \frac{0.059}{2} \log \left( \frac{1.0}{0.01} \right)\]

 

\[E_2 = E^\circ - 0.059 \, \log(100)\]

 

\[E_2 = E^\circ - 0.118 \, \text{V}\]

Since:

• \(E_1 = E^\circ + 0.118 \, \text{V}\)
• \(E_2 = E^\circ - 0.118 \, \text{V}\)

Clearly,

\[E_1 > E_2\]

Hence, the correct answer is \(E_1 > E_2\).