JEE Advance - Chemistry (2018 - Paper 1 Offline - No. 8)
$$\left. {Mg\left( s \right)} \right|M{g^{2 + }}\left( {aq,1\,M} \right)\left\| {C{u^{2 + }}} \right.\left( {aq,1M} \right)\left| {Cu\left( s \right)} \right.$$
the standard $$emf$$ of the cell is $$2.70$$ $$V$$ at $$300$$ $$K.$$ When the concentration of $$M{g^{2 + }}$$ is changed to $$x$$ $$M,$$ the cell potential changes to $$2.67$$ $$V$$ at $$300$$ $$K.$$ The value of $$x$$ is ___________.
(given, $${F \over R} = 11500\,K{V^{ - 1}},$$ where $$F$$ is the Faraday constant and $$R$$ is the gas constant, In $$(10=2.30)$$
Explanation
Equation of cell reaction according to the cell notation given, is
Given, E$$_{cell}^o$$ = 2.70 V, T = 300 K with [Mg2+(aq)] = 1 M and [Cu2+(aq)] = 1 M and n = 2
Further, Ecell = 2.67 V with [Cu2+(aq)] = 1 M and [Mg2+(aq)] = xM and $${F \over R}$$ = 11500 KV$$-$$1 where F = Faraday constant, R = gas constant
From the formula,
Ecell = E$$_{cell}^o$$ $$-$$ $${{RT} \over {nF}}\ln {{[M{g^{2 + }}(aq)]} \over {[C{u^{2 + }}(aq)]}}$$
After putting the given values
$$2.67 = 2.70 - {{RT} \over {2F}}\ln {x \over 1}$$
or $$2.67 = 2.70 - {{R \times 300} \over {2F}} \times \ln x$$
$$ - 0.03 = {{ - R \times 300} \over {2F}} \times \ln x$$
or $$\ln x = {{0.03 \times 2} \over {300}} \times {F \over R}$$
$$ = {{0.03 \times 2 \times 11500} \over {300}} = 2.30$$
So, $$\ln x = 2.30$$
or x = 10 (as given $$\ln (10) = 2.30$$)
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