Part A What is the initial cell potential? Express your answer to two decimal pl

Question

Part A What is the initial cell potential? Express your answer to two decimal places and include the appropriate units. A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/N2+ half-cell at 25 C. The initial concentrations of Ni2 and Zn2+ are 1.50 mol L 1 and 0.130 mol L, respectively Zn2. (aq) + 2 e- Zn(s) Ni2+ (aq) + 2 e- Ni(s) Eo =-0.76 V E o =-0.23 V EeValue Units Submit Part B What is the cell potential when the concentration of Ni2+ has fallen to 0.600 mol L-1? Express your answer to two decimal places and include the appropriate units. a Value Units Submit Part C What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.46 V? Express your answers using two significant figures separated by a comma.

Explanation / Answer

Overall cell reaction,

Ni2+(s) + Zn(s) <==> Ni(s) + Zn2+(aq)

Part A)

Using Nernst relation,

Ecell = Eo - 0.0592/n logQ

Q = [Zn2+]/[Ni2+]

feeding the given values,

Ecell = (-0.23 - (-0.76)) - 0.0592/2 log(0.130/1.50)

         = 0.56 V

Part B)

When [Ni2+] has fallen to = 0.600 M

change in [Ni2+] concentration = 1.50 - 0.60 = 0.90 M

so,

[Zn2+] at this time would be = 0.130 + 0.900 = 1.03 M

Using Nernst relation,

Ecell = Eo - 0.0592/n logQ

feeding the given values,

Ecell = (-0.23 - (-0.76)) - 0.0592/2 log(1.030/0.600)

         = 0.52 V

Part C)

when Ecell = 0.46 V

let x be the change in concentration of [Ni2+] and [Zn2+]

Using Nernst relation,

Ecell = Eo - 0.0592/n logQ

feeding the given values,

0.46 = 0.53 - 0.0592/2 log[(0.130+x)/(1.50-x)]

347.5 - 231.7x = 0.130 + x

x = 347.4/232.7 = 1.493 M

so,

concentration [Ni2+] = 1.5 - 1.493 = 0.0072 M

concentration [Zn2+] = 0.130 + 1.500 = 1.623 M

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