Question

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)

Answer #1

Lets find Eo 1st

from data table:

Eo(Cr2+/Cr(s)) = -0.91 V

Eo(Sn2+/Sn(s)) = -0.13 V

As per given reaction/cell notation,

cathode is (Sn2+/Sn(s))

anode is (Cr2+/Cr(s))

Eocell = Eocathode - Eoanode

= (-0.13) - (-0.91)

= 0.78 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Cr2+]^1/[Sn2+]^1}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Cr2+]^1/[Sn2+]^1}

E = 0.78 - (0.0591/2) log (0.864^1/0.019^1)

E = 0.78-(4.901*10^-2)

E = 0.731 V

Answer: 0.731 V

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

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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 [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 [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

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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 [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 [Mg2 ] = 0.887 M and [Sn2 ] =
0.0150 M. Mg(s)+Sn2+(aq)------->Mg2+ (aq)+Sn(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 ] = 0.771 M and [Ni2 ] =
0.0200 M.

Consider an electrochemical cell, where [Cr2+] = 0.15
M and [Al3+] = 0.0040 M, based on the following
reaction:
3
Cr2+(aq) + 2 Al(s) → 3
Cr(s) + 2 Al3+(aq)
The standard reduction potentials are as follows:
Cr2+(aq) + 2
e- → Cr(s) E° = -0.91 V
Al3+(aq) + 3
e- → Al(s) E° = -1.66 V
What is the cell potential at 25 °C?
Select one:
a. 0.71 V
b. 0.73 V
c. 0.75 V
d. 0.77 V
Can you please hsow working :)

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