Enter your answer in the provided box. what is the emf of a cell consisting of a pb2 / pb half-cell and a pt / h / h2 half-cell

Question

2 years ago

6

if [pb2 ] = 0.57 m, [h ] = 0.090 m and ph2 = 1.0 atm ? v

1 answer:

s2008m [1.1K] · Answer 1 · 2023-07-13T19:01:18+03:00

The emf of a cell is mathematically given as

$E_{cell}$ = 0.0532 v

Generally, the equation for emf is mathematically given as

$Pb/Pb^2^+ //\ H / H_2$

where,

Pb is lead

H is hydrogen

$Pb^{2+}$ / Pb has a standard reduction potential of -0.126 volts, while

$H_2$ / $H_2$ has a standard reduction potential of 0.0 volts.

$E_{cell }= E^0_{cell} - 0.059/n log 1/$ $[Pb^2^+] x [H^+]^2$

$E_{cell}$ = 0.126 - 0.059/2 log 1 / [0.57] x [0.09]^2

$E_{cell}$ = 0.0532 v