For the galvanic cell, Zn(s) + Cu^2+ (0.02 M) \to Zn^2+ (0.04 M) + Cu(s), Ecell = \underline\hspace2cm \times 10 - 2 V. (Nearest integer) [Use : E_Cu/Cu^2 + ^0 = - 0.34 V, E_Zn/Zn^2 + ^0 = + 0.76 V, 2.303RT \over F = 0.059\,V ]

Electrochemistry PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q17 with Solution

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Correct answer: 109

Write the cell reaction and identify electrodes

Given reaction:

Zn(s) + Cu^{2+}(0.02\,M) \rightarrow Zn^{2+}(0.04\,M) + Cu(s)

Zinc is oxidized: $Zn \rightarrow Zn^{2+} + 2e^-$
Copper ion is reduced: $Cu^{2+} + 2e^- \rightarrow Cu$

So, this is a galvanic cell with:

Anode: $Zn/Zn^{2+}$
Cathode: $Cu^{2+}/Cu$

Find standard cell potential

The given standard electrode potentials are in oxidation form:

$E^0_{Cu/Cu^{2+}} = -0.34\,V$
$E^0_{Zn/Zn^{2+}} = +0.76\,V$

Since the actual cathode process is reduction of $Cu^{2+}$ : $E^0_{Cu^{2+}/Cu} = +0.34\,V$

And the anode oxidation is already: $E^0_{Zn/Zn^{2+}} = +0.76\,V$

Thus,

E^0_{cell} = E^0_{oxidation\,(anode)} + E^0_{reduction\,(cathode)}

E^0_{cell} = 0.76 + 0.34 = 1.10\,V

Apply Nernst equation

For the reaction, $n=2$

Reaction quotient:

Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{0.04}{0.02} = 2

Nernst equation:

E_{cell} = E^0_{cell} - \frac{0.059}{n}\log Q

Substitute values:

E_{cell} = 1.10 - \frac{0.059}{2}\log 2

Using $\log 2 \approx 0.3010$ ,

\frac{0.059}{2} = 0.0295

So,

E_{cell} = 1.10 - 0.0295 \times 0.3010

E_{cell} = 1.10 - 0.00888

E_{cell} \approx 1.091\,V

Convert into asked form

They ask:

E_{cell} = \underline{\hspace{2cm}} \times 10^{-2}\,V

Now,

1.091\,V = 109.1 \times 10^{-2}\,V

Nearest integer:

\boxed{109}

Comparison with stored answer

Stored correct answer = $109$

Our derived answer also = $109$

So, the answer agrees.

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View written solutionFree

Correct answer: 109

Write the cell reaction and identify electrodes

Given reaction:

Zn(s) + Cu^{2+}(0.02\,M) \rightarrow Zn^{2+}(0.04\,M) + Cu(s)

Zinc is oxidized: $Zn \rightarrow Zn^{2+} + 2e^-$
Copper ion is reduced: $Cu^{2+} + 2e^- \rightarrow Cu$

So, this is a galvanic cell with:

Anode: $Zn/Zn^{2+}$
Cathode: $Cu^{2+}/Cu$

Find standard cell potential

The given standard electrode potentials are in oxidation form:

$E^0_{Cu/Cu^{2+}} = -0.34\,V$
$E^0_{Zn/Zn^{2+}} = +0.76\,V$

Since the actual cathode process is reduction of $Cu^{2+}$ : $E^0_{Cu^{2+}/Cu} = +0.34\,V$

And the anode oxidation is already: $E^0_{Zn/Zn^{2+}} = +0.76\,V$

Thus,

E^0_{cell} = E^0_{oxidation\,(anode)} + E^0_{reduction\,(cathode)}

E^0_{cell} = 0.76 + 0.34 = 1.10\,V

Apply Nernst equation

For the reaction, $n=2$

Reaction quotient:

Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{0.04}{0.02} = 2

Nernst equation:

E_{cell} = E^0_{cell} - \frac{0.059}{n}\log Q

Substitute values:

E_{cell} = 1.10 - \frac{0.059}{2}\log 2

Using $\log 2 \approx 0.3010$ ,

\frac{0.059}{2} = 0.0295

So,

E_{cell} = 1.10 - 0.0295 \times 0.3010

E_{cell} = 1.10 - 0.00888

E_{cell} \approx 1.091\,V

Convert into asked form

They ask:

E_{cell} = \underline{\hspace{2cm}} \times 10^{-2}\,V

Now,

1.091\,V = 109.1 \times 10^{-2}\,V

Nearest integer:

\boxed{109}

Comparison with stored answer

Stored correct answer = $109$

Our derived answer also = $109$

So, the answer agrees.

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