JEE Advance - Chemistry (2017 - Paper 2 Offline - No. 17)
For the following cell,
$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$
when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]
$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$
when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]
$$1.1F$$
$$2.303RT-2.2F$$
$$2.303RT+1.1F$$
$$-2.2F$$
Explanation
The reaction involved is
$$Zn(s) + C{u^{2 + }}(aq)$$ $$\rightleftharpoons$$ $$Z{n^{2 + }}(aq) + Cu(s)$$
$$\Delta G = \Delta G^\circ + 2.303RT\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}}$$
$$\Delta G = - nFE^\circ + 2.303RT\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}}$$ ........ (1)
Substituting n = 2, E$$^\circ$$ = 1.1 and $$[Z{n^{2 + }}] = 10$$ and $$[C{u^{2 + }}] = 1$$ in Eq. (1), we get
$$\Delta G = ( - 2 \times F \times 1.1) + 2.303RT\log {{10} \over 1}$$
= 2.303 RT $$-$$ 2.2 F
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