Question

when the following skeletal equation is balanced under acidic conditions, what are the coefficients of the species shown? cd + mno2 → cd2+ + mn2+ water appears in the balanced equation as a (reactant, product, neither) with a coefficient of (enter 0 for neither.) which species is the reducing agent?

Explanation:

Step1: Identify oxidation - reduction

In the reaction $Cd + MnO_2
ightarrow Cd^{2+}+Mn^{2+}$, the oxidation state of $Cd$ changes from 0 to + 2 (oxidation), and the oxidation state of $Mn$ in $MnO_2$ (where $Mn$ has an oxidation state of + 4) changes to + 2 (reduction).

Step2: Determine reducing agent

The reducing agent is the species that gets oxidized. Since $Cd$ is oxidized from 0 to + 2, $Cd$ is the reducing agent.

Step3: Balance the redox reaction in acidic medium

First, write the half - reactions:
Oxidation half - reaction: $Cd
ightarrow Cd^{2+}+2e^-$
Reduction half - reaction: $MnO_2 + 4H^++2e^-
ightarrow Mn^{2+}+2H_2O$
Combining the two half - reactions gives the balanced equation: $Cd + MnO_2+4H^+
ightarrow Cd^{2+}+Mn^{2+}+2H_2O$
The coefficient of $Cd$ is 1, the coefficient of $MnO_2$ is 1, the coefficient of $Cd^{2+}$ is 1, the coefficient of $Mn^{2+}$ is 1, and the coefficient of $H_2O$ is 2.

Answer:

Reducing agent: $Cd$
Coefficient of $Cd$: 1
Coefficient of $MnO_2$: 1
Coefficient of $Cd^{2+}$: 1
Coefficient of $Mn^{2+}$: 1
Coefficient of $H_2O$: 2