An electrochemical cell is constructed with an open switch as shown below:

when the switch is closed, mass of tin-electrode increases. If `E^(@) (Sn^(2+)//Sn) = 0.14V` and for `E^(@) (X^(n+)//X) =- 0.78V` and initial emf of the cell is `0.65V`, determine n and indicate the direction of electron flow in the external circuit.

Answer

`n = 3 & X`-electrode to Sn-electrode

Explanation

At cathode:
`Sn^(+2) +2e^(-) rarr Sn`
At anode:
`X rarr X^(+n) +n e^(-)`
Cell reaction
`nSn^(+2) +2X rarr 2X^(+n) +nSn`
`E^(@) =- 0.14 +0.78 = 0.64`
`E = E^(@) -(0.0591)/(2n)log.([X^(+n)]^(2))/([S^(+2)]^(n))`
`0.65 =0.64 -(0.0591)/(2n) [log 0.01 -log (0.5)^(n)]`
`0.01 =- (0.0591)/(2n) [-2 +n log 2]`
`0.03384 n = 2 -0.3010n`
`0.6394 n = 2`
`n = 3.12 =3`