Consider the following electrochemical cell: Cr |Cr(NO_3)_3 (aq)|| AgNO_3(aq)| A

Question

Consider the following electrochemical cell: Cr |Cr(NO_3)_3 (aq)|| AgNO_3(aq)| Ag E_red degree (Ag^+ (aq)/Ag(s)) = +0.80 V E_red degree (Cr^3+ (aq)/Cr(s)) = -0.72 V Calculate the standard cell potential of the cell as written (right - left). Write the balanced half reactions and the overall spontaneous cell reaction. Half reaction on the right (reduction): Half reaction on the left (oxidation): Overall spontaneous cell reaction (balanced): Calculate the equilibrium constant K for the spontaneous cell reaction

Explanation / Answer

a) the standard cell potential of the cell, Eocell is given by:

Eocell = Eocathode - Eoanode

= Eoright - Eoleft

= +0.80 V - (-0.72 V)

Eocell = +1.52 V

b) 1. Half reaction on the right (reduction):

Ag+ (aq) + e-   --------> Ag(s)

2. Half reaction on the left (oxidation):

Cr(s) -----------> Cr3+ (aq) + 3e-

3. Multiplying equation (1.) by 3 to balance the number of electrons on right and left side of the right.and adding it to equation 2.

Overall spontaneous cell reaction (balanced):

3Ag+ (aq) + Cr (s) ---------> 3Ag (s) + Cr3+ (aq)

3. Nernst equation at room temperature = 25oC

Eocell = 0.0592 log K n

where, n = number of electrons lost/gain during the reaction

K = Equilibrium constant

so, log K = n Eocell / 0.0592

For this reaction , n = 3 and Eocell = 1.52 V

log K = (3 X 1.52 ) / 0.0592

log K = 77.027

K = antilog (77.027)

K = 1.06 X 1077

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