Question

1. What is the value (in V) of E^o_cell for the following reaction?

Co²⁺ (aq) + Be (s) → Co (s) + Be²⁺ (aq)

2.What is E^o_cell (in V) for a redox reaction where one electron is transferred with an equilibrium constant (K) of 1.44 x 10^-14?

3.Consider the following reaction:

Cu²⁺ (aq) + Pb (s) → Cu (s) + Pb²⁺ (aq)

What will be E_cell for this reaction (in V) when [Cu²⁺] = 0.500 M and [Pb²⁺] = 0.0350 M, and the temperature is 298 K?

Answer 1

1)

Consider reactions taking place into the cell.

Reaction at anode: Be (s) Be ²⁺ (aq) + 2 e ^-

Reaction at cathode : Co ²⁺ (aq) + 2 e ^- Co (s)

Overall cell reaction :Be (s) + Co ²⁺ (aq)   Be ²⁺ (aq) + Co (s)

The standard emf of the cell is E ⁰ cell = E ⁰ cathode - E ⁰ anode

E ⁰ cell = - 0.282 V - ( - 1.968 V )

E ⁰ cell = 1.686 V

ANSWER : E ⁰ cell = 1.686 V

2)

We have relation, E ⁰ cell = 2.303 R T / n F log K c

E ⁰ cell = ( 2.303 8.314 298 / 1 96487 ) log  1.44 10^-14

E ⁰ cell = - 0.8185 V

ANSWER : E ⁰ cell = - 0.818 V

3)

Consider reactions taking place into the cell

At anode: Pb (s) Pb ²⁺(aq) + 2 e -

At cathode: 2 Cu ²⁺ (aq) + 2 e -    2 Cu(s)

Overall reaction : Cu ²⁺ (aq) + Pb (s) Cu (s) + Pb ²⁺ (aq)

No. of electrons transferred in above reaction = 2

Standard emf of the cell is calculated as, E ⁰ cell = E ⁰ cathode - E ⁰ anode

E ⁰ cell = 0.339 V - ( - 0.126 V )

E ⁰ cell = 0.465 V

We have Nerns't Equation, E cell = E ⁰ cell - [2.303 R T / n F ] log Q

At 25 ⁰ C, 2.303 R T/ F = 0.0591 Therefore, we can write

E cell = E ⁰ cell - 0.0591 / n log [Pb ²⁺  ] / [ Cu ²⁺ ]

= 0.465 - 0.0592 / 2 log 0.0350 / 0.500

= 0.465 - ( - 0.0342 )

= 0.464 + 0.0342

= 0.499 V

ANSWER : E cell = 0.499 V