Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.771 M and [Ni2 ] = 0.0200 M.

Answer #1

Eo ( Ni2+/Ni) = -0.25 V

since Zn is getting oxidised and Ni is getting reduced hence
Zn2+/Zn will be anode and Ni2+/Ni will be cathode

Eo ( cell) = Eo ( cathode) - Eo(anode) = Eo(Ni2+/Ni) - Eo(Zn2+/Zn)
= -0.25 - -0.76 = -0.44 + 0.76 = 0.51 V

now if the conc. of Ni2+ was 1 M and conc. of Zn2+ was 1 M ...then
E (cell) would have been equal to 0.51V .....but since it is not
...hence we have to use the Nernst equation ...

E (cell) = Eo (cell) - RT/nF ln [Zn2+]/[Ni2+]

where R = 8.314 J/K/mole

T = 25 + 273 = 298 K

n = no.of electrons transffered = 2

F = 96500 Coulombs

[Zn2+] = 0.771 M

[Fe2+] = 0.02 M

E (cell) = 0.51 - ( 8.314 X 298) / ( 2 X 96500) X ln
0.771/0.02

E (cell) = 0.51 - 2477.57/193,000 X ln 38.55

E (cell) = 0.51 - 0.0128 X 3.651

E (cell) = 0.51 - 0.046

E (cell) = 0.464 V

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn2 ] = 0.821 M and [Ni2 ] =
0.0200 M. Standard reduction potentials can be found here.
Zn(s) + Ni+2(aq)=Zn+2(aq) + Ni(s)

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn2 ] = 0.752 M and [Sn2 ] =
0.0170 M

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn2 ] = 0.837 M and [Fe2 ] =
0.0100 M. Standard reduction potentials can be found here.
Zn(s)+Fe^2+(aq) <--->Zn^2+(aq)+Fe(s)
E= _______ V

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn 2 ] = 0.768 M and [Sn2 ] =
0.0200 M. Standard reduction potentials can be found here.
Zn(s)+Sn2+(aq)----->Zn2+(aq)+Sn(s)
Zn+2e- ---->Zn = -0.76
Sn+ + 2e- ----> Sn = -0.14

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Cr2 ] = 0.893 M and [Ni2 ] =
0.0130 M. Standard reduction potentials can be found here.
Cr (s) + Ni^2+ (aq) --> <-- Cr^2+ (aq) + Ni (s)

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Mg2 ] = 0.887 M and [Sn2 ] =
0.0150 M. Mg(s)+Sn2+(aq)------->Mg2+ (aq)+Sn(s)

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written at 25.00 °C, given that [Cr2 ] = 0.870 M and [Sn2 ] =
0.0190 M. Standard reduction potentials can be found here.
Cr + Sn2+ <==> Cr2+ + Sn
E= ??V

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Cr2 ] = 0.864 M and [Sn2 ] =
0.0190 M. Standard reduction potentials can be found here.
Cr(s)+Sn^2+(aq) forward and reverse arrow Cr^2+(aq)+Sn(s)

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Mg2 ] = 0.830 M and [Sn2 ] =
0.0140 M. Standard reduction potentials can be found here.
Mg(s) + Sn2+ (aq) <--> Mg2+ (aq) + Sn(s)
Standard... Mg2+(aq) + 2e– → Mg(s.....–2.38
Sn4+(aq) +2e– → Sn2+(aq)....+0.151

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
1.40 Mand 0.130 M , respectively.
What is the initial cell potential?
What is the cell potential when the concentration of Ni2+ has
fallen to 0.600 M ?
What is the concentrations of Ni2+when the cell potential falls
to 0.46 V?
What is the concentration of Zn2+when the cell potential falls
to 0.46 V?

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