Verified Solution • 2025–26 Session moderate

QJEEAdv17Ch1798: JEE Advanced 2017 Chemistry - Class 12 JEEAdv17Ch1798 Solution: For the following cell, when the concentration of is times the…

For the following cell,

Zn(s)∣ZnSO4(aq)∣∣CuSO4(aq)∣Cu(s)Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)

when the concentration of Zn2+Z{n^{2 + }} is 1010 times the concentration of Cu2+,C{u^{2 + }}, the expression for ΔG\Delta G (in J mol−1J\,mo{l^{ - 1}} ) is [ FF is Faraday constant; RR is gas constant; TT is temperature; E0{E^0} (cell) =1.1=1.1 VV ]

Multiple Choice Options

  • A 1.1F1.1F
  • B 2.303RT−2.2F2.303RT-2.2F
  • C 2.303RT+1.1F2.303RT+1.1F
  • D −2.2F-2.2F

Step-by-Step Verified Solution

The reaction involved is

Zn(s)+Cu2+(aq)Zn(s) + C{u^{2 + }}(aq) ⇌\rightleftharpoons Zn2+(aq)+Cu(s)Z{n^{2 + }}(aq) + Cu(s)

ΔG=ΔG∘+2.303RTlog⁡[Zn2+][Cu2+]\Delta G = \Delta G^\circ + 2.303RT\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}}

ΔG=−nFE∘+2.303RTlog⁡[Zn2+][Cu2+]\Delta G = - nFE^\circ + 2.303RT\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}} ........ (1)

Substituting n = 2, E ∘^\circ = 1.1 and [Zn2+]=10[Z{n^{2 + }}] = 10 and [Cu2+]=1[C{u^{2 + }}] = 1 in Eq. (1), we get

ΔG=(−2×F×1.1)+2.303RTlog⁡101\Delta G = ( - 2 \times F \times 1.1) + 2.303RT\log {{10} \over 1}

= 2.303 RT −- 2.2 F

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