Question

What is Eo for the cell Ag l AgBr(s) I Br- (a1), Fe3+ (a2), Fe2+ (a3) l Pt . Given : Standard half-cell reduction potentials as
AgBr l Ag, (Eo)1/2 = 0.0713 V, Fe3+ l Fe2+, (Eo)1/2 = 0.771 V.
If a1 = 0.34, a2 = 0.1 & a3 = 0.02, then find Q, the reaction Quotient is Q = 0.588. Hence find E for the cell.

Answer 1

Question #50893, Chemistry, Physical Chemistry

What is Eo for the cell Ag | AgBr(s) | Br⁻ (a1), Fe³⁺ (a2), Fe²⁺ (a3) | Pt. Given: Standard half-cell reduction potentials as

AgBr | Ag, (Eo)1/2 = 0.0713 V, Fe³⁺ | Fe²⁺, (Eo)1/2 = 0.771 V.

If a1 = 0.34, a2 = 0.1 & a3 = 0.02, then find Q, the reaction Quotient is Q = 0.588. Hence find E for the cell.

Answer:

$$E = E _ {0} + \frac {0 . 5 9 2}{n} \lg \left(a _ {\mathrm {o x}} / a _ {\mathrm {r e d}}\right)$$ $$E _ {1} = 0. 0 7 1 3 + 0. 5 9 2 ^ {*} \lg (0. 3 4) = - 0. 2 \mathrm {V} \quad n = 1$$ $$E _ {2} = 0. 7 7 1 + 0. 5 9 2 ^ {*} \lg (0. 0 2 / 0. 1) = 0. 3 5 7 \mathrm {V} \quad n = 1$$ $$E = E _ {1} - E _ {2} = - 0. 2 - 0. 3 5 7 = - 0. 5 5 7 \mathrm {V}$$

https://www.AssignmentExpert.com