Calculate E cell for each of the following balanced redox reactions. Part A O2(g

Question

Calculate Ecell for each of the following balanced redox reactions.

Part A

O2(g)+2H2O(l)+4Ag(s) 4OH(aq)+4Ag+(aq)

Express your answer using two significant figures.

Part B

Br2(l) + 2I(aq) 2Br(aq) + I2(s)

Express your answer using two significant figures.

Part C

PbO2(s)+4H+(aq)+Sn(s)Pb2+(aq)+2H2O(l)+Sn2+(aq)

Express your answer using two significant figures.

Part D

Which of the reactions are spontaneous as written.

Check all that apply.

Check all that apply.

Part B

Br2(l) + 2I(aq) 2Br(aq) + I2(s)

Express your answer using two significant figures.

Part C

PbO2(s)+4H+(aq)+Sn(s)Pb2+(aq)+2H2O(l)+Sn2+(aq)

Express your answer using two significant figures.

Part D

Which of the reactions are spontaneous as written.

Check all that apply.

Check all that apply.

PbO2(s)+4H+(aq)+Sn(s)Pb2+(aq)+H2O(l)+Sn2+(aq) O2(g)+2H2O(l)+Ag(s)4OH(aq)+4Ag+(aq) Br2(l)+2I(aq)2Br(aq)+I2(s)

Explanation / Answer

Part A:

The half reactions are:

O2 + 2H2O + 4e- ----> 4OH- E0= 0.40V (cathode)

(Ag ----> Ag+ + e-) x4 E0= 0.80V (anode)

-----------------------------------

O2(g) + 2H2O(l) + 4Ag(s) 4OH(aq) + 4Ag+(aq)

E0cell= E0cathode - E0anode= 0.40- (0.80)= -0.40V

Part B:

Tha half reactions are:

Br2 + 2e- ----> 2Br-   E0=1.07V (cathode)

2I- ----> I2 + 2e-   E0= 0.53V (anode)

-----------------------

Br2(l) + 2I(aq) 2Br(aq) + I2(s)

E0cell= E0cathode - E0anode= 1.07 - 0.53 = 0.54V

Part C:

The half reactions are:

PbO2 + 4H+ + 2e- -----> Pb+2 + 2H2O E0= 1.70V (cathode)

Sn -----> Sn+2 + 2e- E0= -0.14V (anode)

-----------------------------------------------------

PbO2(s)+4H+(aq)+Sn(s)Pb2+(aq)+2H2O(l)+Sn2+(aq)

E0cell= 1.70 -(-0.14)= 1.84V

Part D:

A reaction is spontaneous when the E0cell is positive.

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