A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 C.

Question

A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 C. The initial concentrations of Ni2+ and Zn2+ are 1.60 M and 0.130 M , respectively

A.) What is the initial cell potential? Express your answer using two significant figures

B.) What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M ? Express your answer using two significant figures

C.) What is the concentrations of Ni2+ when the cell potential falls to 0.45 V ? Express your answer using one significant figure

D.) What is the concentration of Zn2+ when the cell potential falls to 0.45 V ? Express your answer using two significant figures.

Explanation / Answer

Zn----> Zn+2 +2e-   Eo=0.76V

Ni+2+2e- -------> Ni Eo =-0.25 V

Addition of the above equations give

Zn+ Ni+2 ---------> Zn+2 +Ni Eo= 0.51V

E= EO- 0.0591/2*log [Zn+2]/ [Ni+2]

=0.51-0.0591/2*log (0.130/1.60) =0.542215 V

when the concentration of Ni+2 drops to 0.5M,

E =0.51-0.0591/2*log (0.130/0.5)=0.527288V

c) 0.45= 0.51- 0.0591/2 *log (0.130/Ni+2)

0.02955 *log (0.130/Ni+2)= 0.06

log (0.130/Ni+2)= 0.06/0.02955 =2.03

0.130/Ni+2 =107

Ni+2 =0.130/107 = 0.001215

since cell potentital is at 0.45V

Zn+2/1.6= 107

Zn+2 =107*1.6=171.2 M

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