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In an electrochemical reaction of lead, at standard temperature, if E Pb 2 + / Pb o = m Volt and E Pb 4 + / Pb o = n Volt , then the value of E ° Pb 2 + / Pb 4 + is given by m – xn . The value of x is _______. (Nearest integer)

Correct answer

0

Step-by-step solution

The relation between Gibbs Free energy and emf of the cell can be related as follows, ∆ G = - nFE cell The electrode reactions can be written as follows, Pb 2 + + 2 e - → Pb               ∆ G 1 = - 2 FE Pb 2 + / Pb o Pb 4 + + 4 e - → Pb               ∆ G 2 = - 4 FE Pb 4 + / Pb o Pb 4 + + 2 e - → Pb 2 +               ∆ G 3 = - 2 FE Pb 4 + / Pb 2 + o Now, ∆ G 3 = &#8

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