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Question

Calculate the standard potential for the reaction,
Hg2Cl2+Cl22Hg++4Cl
Given:
Hg2Cl2+2e2Hg+2Cl;E=0.270 volt
Hg2+22Hg2++2e;E=0.92 volt
2HgHg2+2+2e;E=0.79 volt
Cl2+2e2Cl;E=1.36 volt.

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Solution

We can write all the equations with their G values.
Hg2Cl2+2e2Hg+2Cl;G=nFEcell=52110J(1)
Hg222Hg2++2e;G=177560J(2)
2HgHg2+2+2e;]triangleG=152470J(3)
Cl22e2Cl;G=262480J(4)
Adding reaction (2) and (3) we get,
2Hg2Hg2++4e;G=330030(5)
Adding reaction (1) and (4), we get,
Hg2Cl2+Cl2+4e4Cl+2Hg;G=314590(6)
Adding reactionn (5) and (6), we get,
Hg2Cl2+Cl22Hg++4Cl;G=540400(7)
Ecell=GnF=154404×96500=0.04V

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