Question

Given:
(i) Cu²⁺ + 2e⁻ ⟶ Cu, E⁰ = 0.337 V
(ii) Cu²⁺ + e⁻ ⟶ Cu⁺, E⁰ = 0.153 V
Electrode potential, Eo for the reaction,
Cu⁺ + e⁻ ⟶ Cu, will be :
(a) 0.90 V (b) 0.30 V
(c) 0.38 V (d) 0.52 V

Answer 1

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${Cu}^{2+}+2{e}^{-} \longrightarrow Cu -------(1){E}_{1}$

${Cu}^{2+}+{e}^{-} \longrightarrow {Cu}^{+} -------(1){E}_{2}$

Net equation

${Cu}^{+}+{e}^{-} \longrightarrow Cu-------(3)$

$E _{3} = \frac{ n_{1}(E_{1}) - n_{2}(E_{2}) }{ n_{3}} \\ E_{3} = \frac{2 \times 0.337 - 1 \times 0.153}{1} \\ E_{3} = 0.521 \: V \\ \\ Hence, \: option \: D \: is \: correct.$

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Answer 2

Hello Friend..!!

The answer of ur question is..!!

Option.D

Thank you..!!