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The EMF of the cell M|M^(n+)(0.02M) ||H^...

The EMF of the cell `M|M^(n+)(0.02M) ||H^(+)(1M)|H_(2)(g) (1atm)Pt` at `25^(@)C` is `0.81V`. Calculate the valency of the metal if the standard oxidation potential of the metal is `0.76V`.

Text Solution

Verified by Experts

The correct Answer is:
`n = 2`
`E_(cell) = (E_(cell)^(@) - (0.0591)/(n) log.([2 xx 10^(-2)])/([H^(+)]))`
`M rarr M^(n+) +n e^(-)`
`(n)/(2) (2H^(+) +2e^(-) rarr H_(2))`
`0.81 = 0.76 -(0.0591)/(n) log 2 xx 10^(-2)`
`n =2`
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