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Question: $2MnO_4^- + b \ C_2O_4^{2-} + c \ H^+ \rightarrow x \ Mn^{2+} + y \ CO_2 + z \ H_2O$ If the above e...

2MnO4−+b C2O42−+c H+→x Mn2++y CO2+z H2O2MnO_4^- + b \ C_2O_4^{2-} + c \ H^+ \rightarrow x \ Mn^{2+} + y \ CO_2 + z \ H_2O

If the above equation is balanced with integer coefficients, the value of c is ........ (Round off to the Nearest Integer)

Answer

16

Explanation

Solution

1. Identify Half-Reactions:

  • Reduction: MnO4−MnO_4^- (Mn oxidation state +7) is reduced to Mn2+Mn^{2+} (Mn oxidation state +2).
  • Oxidation: C2O42−C_2O_4^{2-} (C oxidation state +3) is oxidized to CO2CO_2 (C oxidation state +4).

2. Balance Half-Reactions:

  • Reduction: MnO4−→Mn2+MnO_4^- \rightarrow Mn^{2+} Balance O by adding H2OH_2O: MnO4−→Mn2++4H2OMnO_4^- \rightarrow Mn^{2+} + 4H_2O Balance H by adding H+H^+: MnO4−+8H+→Mn2++4H2OMnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O Balance charge by adding electrons (e−e^-): MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O (Charge: Left = -1+8 = +7; Right = +2)

  • Oxidation: C2O42−→CO2C_2O_4^{2-} \rightarrow CO_2 Balance C: C2O42−→2CO2C_2O_4^{2-} \rightarrow 2CO_2 Balance charge by adding electrons (e−e^-): C2O42−→2CO2+2e−C_2O_4^{2-} \rightarrow 2CO_2 + 2e^- (Charge: Left = -2; Right = 0)

3. Equalize Electrons and Combine: To balance the electrons, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5:

  • 2×(MnO4−+8H++5e−→Mn2++4H2O)  ⟹  2MnO4−+16H++10e−→2Mn2++8H2O2 \times (MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O) \implies 2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O
  • 5×(C2O42−→2CO2+2e−)  ⟹  5C2O42−→10CO2+10e−5 \times (C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-) \implies 5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-

Add the two balanced half-reactions: (2MnO4−+16H++10e−)+(5C2O42−)→(2Mn2++8H2O)+(10CO2+10e−)(2MnO_4^- + 16H^+ + 10e^-) + (5C_2O_4^{2-}) \rightarrow (2Mn^{2+} + 8H_2O) + (10CO_2 + 10e^-) Cancel electrons: 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O

4. Identify Coefficient 'c': The balanced equation is 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O. Comparing this with the given equation 2MnO4−+b C2O42−+c H+→x Mn2++y CO2+z H2O2MnO_4^- + b \ C_2O_4^{2-} + c \ H^+ \rightarrow x \ Mn^{2+} + y \ CO_2 + z \ H_2O, we find that the coefficient of H+H^+ is c=16c = 16.

5. Rounding: The question asks to round off the value of c to the nearest integer. Since c=16c=16 is already an integer, the rounded value is 16.