Question

The emf of the following cell Cd/Cd2+(0.002 M)// Cd2+ (X M) /Cd is 0.061 V at 25 oC. Find the value of ‘X’. If water is added to anode, what happens to cell emf.​

Answer 1

the cell were decompose by water

Answer 2

Given:

Emf of the following cell Cd/Cd⁺²//Cd⁺²/Cd is 0.0061V at 25°C.

To find:

Concentration of Cd⁺² ions = XM

Solution:

Nernst equation used,

${{\text{E}}}}} = {{\text{E}}^{\text{0}}} - \left( {\frac{{R \cdot T}}{{n \cdot F}}} \right) \cdot \ln \left( {\frac{{\prod {{{\left( {{c_{{\text{Red}}}}} \right)}}} }}{{\prod {{{\left( {{c_{Ox}}} \right)}} }}} \right)}$

here, E = emf of the cell (i.e. given 0.061V)

         E⁰ = standard emf of the cell ( 0V)

         R = Gas constant (8.314J mol⁻¹ K⁻¹)

         c(red) = concentration on cathode (i.e.XM)

         c(oxi) = conentration on anode (i.e. 0.002M)

         n  = no. of electrons in a balanced redox reaction

         T = temperature(298K)

         F = faraday constant

now, putting all the values we get,

${{\text{0.061}}}}} = {{\text{0}}^{\text{}}} - \left( {\frac{{8.314 \cdot 298 \cdot2.303}}{{2 \cdot 95400}}} \right) \cdot \log \left( {\frac{{\prod {{{\left( {{X_{{\text{}}}}} \right)}}} }}{{\prod {{{\left( {{0.002_{}}} \right)}}} }}} \right)}$

by solving this we get

X = antilog(-2.0608)×0.002

X = 0.0087×0.002

X = 0.000017M

So, the concentration of Cd⁺² is 0.000017M

Addition of water to anode will going to decrease the concentration of ions present on anode.