O2(g)+2h2o(l)+4ag(s) → 4oh−(aq)+4ag+(aq)

Calculate E∘cell for each of the following balanced redox reactions. Part A: O2(g)+2H2O(l)+4Ag(s) → 4OH−(aq)+4Ag+(aq) Part B: Br2(l) + 2I−(aq) → 2Br−(aq) + I2(s) Part C: PbO2(s)+4H+(aq)+Sn(s)→Pb2+(aq)+2H2O(l)+Sn2+(aq) Part D: Which of the reactions are spontaneous as written. Check all that apply

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

Answer

General guidance

Concepts and reason
Write the half-cell reactions for the given balanced reactions. Calculate the by using the standard cell potential of the given half cells. Check the of the reactions to decide whether the reaction is spontaneous or not.

Fundamentals

Electrochemical cell is a chemical cell which generates electrical energy from the chemical reaction taking place in the reaction. It consists of cathode and anode. Oxidation is defined as the process in which a substance loses electrons and gets oxidized whereas reduction is defined as the process in which the substance accepts electrons and gets reduced. Cathode is defined as electrode through which the electrons enter the cell. It is negatively charged electrode. In this electrode, the reduction process takes place. Anode is defined as electrode through which the electrons leave the cell. It is positively charged electrode. In this electrode, the oxidation process takes place. The of the reaction is calculated using the following formula: Ecode = Ecathode - En mode Here, chode is the standard cell potential of the cathode and is the standard cell potential of the anode. A reaction is said to be spontaneous if the of the reaction is positive.

Step-by-step

Step 1 of 7

A The given reaction is as follows: 02(g)+2H20 (1) +4Ag(s) →40H (aq)+4Ag* (aq) The half-cell reactions are written as follows: At anode(oxidation): Ag(s) → Ag+ (aq)+e At cathode(reduction): 0,(g)+2H,0(1)+4e →40H (aq)

The complete cell reaction is written by incorporating both oxidation and reduction reactions. The species being oxidized is and the species being reduced is . So, is written at anode and is written at anode. The species at anode loses electron whereas the species at cathode gains electrons.

Step 2 of 7

The standard cell potential is calculated as follows: Ecev = Ecathocke - Emede
= 0.40 V-0.80 V
= -0.40 V

Part A

The for the given reaction is |-0.40 V.


The standard values of cell potential are as follows: Substitute the values in the standard cell potential formula to calculate the .

Step 3 of 7

B The given reaction is as follows: Br_(1)+21+ (aq) + 2Br (aq) +1,(s) The half-cell reactions are written as follows: At anode(oxidation): 21+ (aq) +1,($)+2e At cathode(reduction): Brz (1)+2e → 2Br (aq)

The complete cell reaction is written by incorporating both oxidation and reduction reactions. The species being oxidized is and the species being reduced is . So, is written at anode and is written at anode. The species at anode loses electron whereas the species at cathode gains electrons.

Step 4 of 7

The standard cell potential is calculated as follows: Ecode = Ecathode - Emode
= 1.09 V-0.54 V
= 0.55 V

Part B

The for the given reaction is 0.55 V.


The standard values of cell potential are as follows: Substitute the values in the standard cell potential formula to calculate the .

Step 5 of 7

C The given reaction is as follows: PbO, (s)+4H* (aq) +Sn(s) → Pb2+ (aq)+24,0(1)+Snº* (aq) The half-cell reactions are written as follows: At anode(oxidation): Sn(s)
Sn2+ (aq) +2e At cathode(reduction): PbO2 (s)+4H* (aq)+2e → Pb2+ (aq)+2H,0(1)

The complete cell reaction is written by incorporating both oxidation and reduction reactions. The species being oxidized is and the species being reduced is . So, is written at anode and is written at cathode. The species at anode loses electron whereas the species at cathode gains electrons.

Step 6 of 7

The standard cell potential is calculated as follows: Ecd= Eethode - Emock
= 1.46V-(-0.4V)
=1.60 V

Part C

The for the given reaction is 1.60V.


The standard values of cell potential are as follows: Substitute the values in the standard cell potential formula to calculate the .

Step 7 of 7

D The spontaneous reactions are as follows: PbO, (s)+4H* (aq) +Sn(s) → Pb2+ (aq)+24,0(1)+Snº* (aq) Br_(1)+21+ (aq) + 2Br (aq) +1,(s)

Part D

The spontaneous reactions are as follows: PbO, (s)+4H* (aq) +Sn(s) → Pb2+ (aq)+24,0(1)+Snº* (aq) Br_(1)+21+ (aq) + 2Br (aq) +1,(s)


A spontaneous reaction is the one with the positive value of of the reaction.

Answer

Part A

The for the given reaction is |-0.40 V.

Part B

The for the given reaction is 0.55 V.

Part C

The for the given reaction is 1.60V.

Part D

The spontaneous reactions are as follows: PbO, (s)+4H* (aq) +Sn(s) → Pb2+ (aq)+24,0(1)+Snº* (aq) Br_(1)+21+ (aq) + 2Br (aq) +1,(s)

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