Question

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

Answer #1

**we know that**

**oxidation takes place at anode**

**and**

**reduction takes place at cathode**

**so**

**oxidation : anode**

**Mg ----> Mg+2**

**reduction : cathode**

**Sn+2 ----> Sn**

**now**

**Eo cell = Eo cathode - Eo anode**

**Eo cell = 0.151 + 2.38**

**Eo cell = 2.531**

**now**

**consider the cell equation**

**Mg (s) + Sn+2 ----> Mg+2 + Sn (s)**

**the reaction quotient is given by**

**Q = [Mg+]2 / [Sn+2]**

**according to nernst equation**

**E = Eo - ( 0.05916 / n) log Q**

**E = Eo - ( 0.05916 / n) log [Mg+2] / [Sn+2]**

**here**

**n = 2 as two electrons are transferred**

**so**

**E = 2.531 - ( 0.05916 / 2) log [ 0.83 /
0.014]**

**E = 2.47855**

**so**

**the cell potential is 2.47855 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 [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 [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 [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 [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 [Zn2 ] = 0.752 M and [Sn2 ] =
0.0170 M

1. What is the calculated value of the cell potential at 298K
for an electrochemical cell with the following reaction, when the
Pb2+ concentration is
4.09×10-4 M and the
Mg2+ concentration is
1.03 M ?
Pb2+(aq) +
Mg(s) --->
Pb(s) +
Mg2+(aq)
Answer: ___V
The cell reaction as written above is spontaneous for the
concentrations given____. (true or false)
(From the table of standard reduction potentials: )
Pb2+(aq) + 2 e- --> Pb(s)
-0.126
Mg2+(aq) + 2 e- --> Mg(s)...

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.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 9 minutes ago

asked 28 minutes ago

asked 29 minutes ago

asked 45 minutes ago

asked 54 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago