`underset((1atm25^@C))(Pt:H_2)//underset((0.01M))(HCOOK)//// underset((0.01 M))((NH_4 Cl)) // underset((1 atm ,25^@ C))(H_2,Pt )K_b` value of `HCOO^(-) and NH_3 ` are ` 10^(-6) and 10^(-8)` respectively. The potential of the given cell is

Options

A. `-0.0591 V `
B. `0.3546 V `
C. ` 0.2364 V `
D. `-0.3564 V `

Answer

B

Explanation

`[HCOOK] = 10^(-2) implies (H^(oplus))_(a) = sqrt((K_w xx K_a)/(C)) , [H^(oplus)]_a = sqrt((10^(-14) xx 10^(-8))/(10^(-2))) = ((10^(-7) xx 10^(-4))/(10^(-1)))`
`(H^(oplus))_a = 10^(-11 + 1) = 10^(-10) and (NH_4Cl) = 10^(-2)`
` (H^(oplus))_(C) = sqrt((K_w xx C )/(K_b)) = sqrt((10^(-14) xx 10^(-2))/(10^(-8))) , (H^(oplus))_(C) = (10^(-7) xx 10^(-1))/(10^(-4)) = 10^(-8 + 4) = 10^(-4) and E = (+0.0591)/(1) log""([H^(oplus)]_c)/([H^(oplus)]_a) = (+ 0.0591)/(1) log ""((10^(-4))/(10^(-10))) = 0.0591 xx 6 = 0.3546V `