if hydrogen electrode is dipped in 2 solutions of pH = 4 and pH = 6 at 1 atm pressure and salt bridge is connected, the e.m.f. of resulting cell is ……V

Answer

0.12

Explanation

pH = 4 `[H^(+)] = 10^(-4) M `
pH =6 `[H^(+) ] = 10^(-6) ` M
The, electrode which has lower `[H^(+)]` acts as anode while electrode with higher `[H^(+)]` acts as cathode.
Hence,
Anode reaction:
`H_(2) (P_(1)) rarr 2 H^(+) (10^(-6) M) +2e^(-) `
Cathode reaction:
`2H^(+) (10^(-4) M) + 2e^(-) rarr H_(2) (P_(2))`
`E_("anode") = - (RT )/(2 F) "In" ([H^(+)]^(2))/(P_(1))`
`= - (0.0591)/(2) "log" ((10^(-6))^(2))/(1)`
`E_("cathade ") = - (RT )/(2F ) " In " ( P_(2))/([H^(+)]^(2))`
`- (0.0591) /(2) "log" (1)/((10^(-4))^(2))`
`E_("cell") = E_("anode") + E_("cathode") `
`= - (0.0591)/(2) "log" ((10^(-6))^(2))/(1) - (-0.0591)/(2) "log" (1)/((10^(-4))^(2))`
`E_("cell") = - (0.0591)/(2) "log" ((10^(-6))^(2))/((10^(-4))^(2))`
`= - 0.591 xx "log" 10^(-2)`
`= - 0.0591 xx (-2) = 0.12 V`