For the reaction Zn(s) + Cu 2+ (aq) ⇌ Cu(s) + Zn 2+ (aq) Reaction Quotient = [ Zn ^ 2+ ]…

Chemistry · JEE Advanced · NTA Exams — Electrochemistry

For the reaction
Zn(s) + Cu²⁺ (aq) \(\text { ⇌ }\) Cu(s) + Zn²⁺(aq)
Reaction Quotient = \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\), variation of E_cell with Q is given by
(where Q = concentration quotient)

OA = 1.10 volts, hence
When E_cell is 1.1591 volts. It implies,

\(\frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Zn}^{2+}\right]}=0.01\)
\(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=0.01\)
\(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=0.1\)
\(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=1\)

Answer

(B) [ Zn ^ 2+ ] [ Cu ^ 2+ ] =0.01

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For the reaction Zn(s) + Cu 2+ (aq) ⇌ Cu(s) + Zn 2+ (aq) Reaction Quotient = [ Zn ^ 2+ ]…

Answer

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