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pH is proportional to the cell potential and it is indicated by the Nernst equation
According to Nernst equation -if the reaction occurs in preence of acidic media - Ecell = E0(cell)
- 0.0592/n log 1/[H+] [because all other species are at standard conditions] Therefore - Ecell =
0.42 V - (0.0592/n) log 1/[H+] Let n = 1, then - 0 V = 0.42 V - 0.592 pH 0.0592 pH = 0.42
Therefore - pH = 0.42/0.0592 = 6.75
Solution
pH is proportional to the cell potential and it is indicated by the Nernst equation
According to Nernst equation -if the reaction occurs in preence of acidic media - Ecell = E0(cell)
- 0.0592/n log 1/[H+] [because all other species are at standard conditions] Therefore - Ecell =
0.42 V - (0.0592/n) log 1/[H+] Let n = 1, then - 0 V = 0.42 V - 0.592 pH 0.0592 pH = 0.42
Therefore - pH = 0.42/0.0592 = 6.75

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pH is proportional to the cell potential and it i.pdf

  • 1. pH is proportional to the cell potential and it is indicated by the Nernst equation According to Nernst equation -if the reaction occurs in preence of acidic media - Ecell = E0(cell) - 0.0592/n log 1/[H+] [because all other species are at standard conditions] Therefore - Ecell = 0.42 V - (0.0592/n) log 1/[H+] Let n = 1, then - 0 V = 0.42 V - 0.592 pH 0.0592 pH = 0.42 Therefore - pH = 0.42/0.0592 = 6.75 Solution pH is proportional to the cell potential and it is indicated by the Nernst equation According to Nernst equation -if the reaction occurs in preence of acidic media - Ecell = E0(cell) - 0.0592/n log 1/[H+] [because all other species are at standard conditions] Therefore - Ecell = 0.42 V - (0.0592/n) log 1/[H+] Let n = 1, then - 0 V = 0.42 V - 0.592 pH 0.0592 pH = 0.42 Therefore - pH = 0.42/0.0592 = 6.75