For the redox reaction- Zn-Cu2-0-1M- Zn2-1M-Cu taking place in a cell E-cell-1-10V- E-Cell for the galvanic cell will be-

The correct option is C 1.07 VBY Nernst's equation, _c ∈ 11=E_cell ^0 0.0591nlogθ_1 (at 5^∘C Here, theta=[z n^+2][(μ^+2].=10 · 1=10 So, ...

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Standard XII

Chemistry

Types of Redox Reactions

For the redox...

Question

For the redox reaction, $Z n + C u_{(0.1 M)}^{2 +} \to Z n_{(1 M)}^{2 +} + C u$ taking place in a cell $E_{c e l l}^{⊖} = 1.10$ V. $E_{C e l l}^{⊖}$ for the galvanic cell will be?

2.14

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1.80

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1.07

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0.82

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Solution

The correct option is C $1.07$ V
BY Nernst's equation, $_{c \in 11} = E_{cell}^{0} - \frac{0.0591}{n} log θ_{1}$ (at $5^{\circ} C$

Here, $t h e t a = \frac{[z n^{+ 2}]}{[(μ^{+ 2}]} = \frac{1}{0 \cdot 1} = 10$

So, $_{cell} = 1.10 - \frac{0.0591}{2} log 10 . = 1.10 - 0.0295$
$= 1.0705 V$
Thus, option (C) is correct.

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Similar questions

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For the redox reaction, Zn+Cu2+(0.1M)→Zn2+(1M)+Cu taking place in a cell E⊖cell=1.10V. E⊖Cell for the galvanic cell will be?

The correct option is C 1.07VBY Nernst's equation, c∈11=E0cell −0.0591nlogθ1 (at 5∘CHere, theta=[zn+2][(μ+2]=10⋅1=10So, cell =1.10−0.05912log10.=1.10−0.0295=1.0705 VThus, option (C) is correct.

For the redox reaction, $Z n + C u_{(0.1 M)}^{2 +} \to Z n_{(1 M)}^{2 +} + C u$ taking place in a cell $E_{c e l l}^{⊖} = 1.10$ V. $E_{C e l l}^{⊖}$ for the galvanic cell will be?