A voltaic cell consists of two half-cells. One half-cell contains a chromium ele

Question

A voltaic cell consists of two half-cells. One half-cell contains a chromium electrode immersed in 1.00 M Cr(NO3)3 solution. The second half-cell contains a cobalt electrode immersed in 1.00 M Co(NO3)2 solution. Cobalt plates out on the cobalt electrode as the voltaic cell runs. The beginning voltage of the cell is +0.467 V at 25°C. The standard electrode potential (standard reduction potential) of chromium at 25°C is 0.744 V.

(a) Write a balanced half-reaction equation for the reaction occurring at the anode and a second balanced equation for the reaction occurring at the cathode. (Include states-of-matter under the given conditions in your answer.)

(b) Write a balanced equation to show the net reaction for the cell. (Include states-of-matter under the given conditions in your answer.)

(c) Determine the standard electrode potential for cobalt.

Explanation / Answer

a) Oxidation half -rxn: (at anode)

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

Reduction half-rxn: (at cathode)

Co2+(aq) +2e --->Co(s)

B) Cr(s) ---->Cr3+(aq)+3e ]*2

Co2+(aq) +2e --->Co(s) ]*3

2Cr(s) +3Co2+(aq)----------->2Cr3+(aq)+3 Co(s) [net cell rxn]

C) using Nerst equation,

Ecell=Eocell-RT/nF ln Q

Where Q=rxn quotient=[Cr3+]^2/[Co2+]^3

n=mol of electron exchanged=6

R=universal gas constant=8.314J/Kmol

F=faraday’s constant=96485 C/mol

T=298K

So,Ecell=Eocell-0.0591V/6 *log [Cr3+]^2/[Co2+]^3=Eocell-0.0591V/6 *log (1M)^2/[1M)^3=Eocell

Or, Ecell=Eocell=+0.467V

Also Eocell=Eo(red)cathode-Eo(red)anode=Eo(red,Co)-Eo(red,Cr)

Or,0.467V=Eo(red,Co)-(-0.744V)

Eo(red,Co)=0.467-0.744=-0.277V

Answr=-0.277V

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