Question

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

Answer #1

From the reaction given it is clear that zinc is oxidized i.e. electron are losed by zinc . Iron is reduced i.e. it gains electrons

From the reaction above lets write half reactions

Fe^{+2}_{(aq)} + 2e^{-} -------> Fe
_{(s)} E°_{red} = -0.45

Zn_{(s)} -------> Zn^{2+}_{(aq)} +
2e^{-} E°_{oxd} = +0.76

Net reaction - Zn_{(s)} + Fe^{+2}_{(aq)}
-----> Zn^{+2}_{(aq)} + Fe_{(s)}
E°_{cell} = 0.31

Now will we use Nernst equation

E_{cell}= E^{°}_{cell} - (RT/nF) ln
Q

E_{cell} = E^{°}_{cell} - (RT/nF) ln
([Zn^{2+}] / [Fe^{2+}])

Here

E^{°}_{cell} = 0.31

R = 8.314 ( gas constant )

n = no. Of electrons transferred = 2

F = 96500 columbs

[Zn^{2+}] = 0.837 M

[Fe^{2+}] = 0.0100 M

T = 25^{°} c = 273 +25 = 298 k

Substitute

E_{cell} = 0.31 - ((8.314*298)/(2*96500)) ln
(0.837/0.0100)

= 0.31 - (2477.572/193000) ln (87)

= 0.31 - 2.303 * 0.0128 * log (87) (2.303 is multiplied to remove the natural log ln )

= 0.31 - 2.303 * 0.0128 * 1.9395

= 0.31 - 0.0571

= 0.2529 V

Hence E = 0.2529 V

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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 [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 [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 [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 [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

Calculate the cell potential for the following reaction as
written at 25.00 °C, given that [Zn2 ] = 0.752 M and [Sn2 ] =
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written at 25.00 °C, given that [Zn2 ] = 0.771 M and [Ni2 ] =
0.0200 M.

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)

The following cell has a potential of 0.040 V at 25°C.
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Zn2+(aq) + 2e− → Zn(aq) Eo = − 0.760 Cr3+(aq) + 3e− → Cr(s) V Eo =
− 0.740 V

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