Cu⊕+e−→Cu,E⊖=x1 volt Cu2++2e−→Cu,E⊖=x2 volt, then for Cu2++e−→Cu,E⊖ (volt) will be:
A
x1−2x2
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B
x1+2x2
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C
x1−x2
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D
2x1−x1
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Solution
The correct option is C2x1−x1 Cu⊕+e−→Cu;E⊖1=x1 V Cu2++2e−→Cu;E⊖2=x2 V Net equation Cu2++e−→Cu⊕;E3=? is obtained by equations (ii) - (i) ∴E3=n2E2−n1E1n3 =2×x2−1×x11=2x2−x1