Write a balanced molecular equation for the reaction between KMnO_4 and Kl in ba

Question

Write a balanced molecular equation for the reaction between KMnO_4 and Kl in basic solution. The skeleton ionic reaction is: MnO_4^-(aq) + l-(aq) rightarrow MnO_4^2- (aq) + IO_3^-(aq) 2. In one compartment of a voltaic cell, a graphite rod dips into an acidic solution of K_2Cr_2O_7 and Cr(NO_3)_3; in the other compartment, a tin bar dips into a Sn(NO_3)_2 solution. A KNO_3 salt bridge joins them. The tin electrode is negative relative to the graphite. Diagram the cell and write the balanced equations and the cell notation. 3. A voltaic cell based on the reaction between aqueous Br_2 and vanadium(III) ions has Edegree_cell = 1.39 V: Br_2(aq) + 2V^3+(aq) + 2H_2O(l) rightarrow 2VO_2^+ (aq) + 4H^+(aq) + 2Br^-(aq) What is Edegree_vanadium, the standard electrode potential for the reduction of VO_2+ to V^3+? 4. Is the following reaction spontaneous as written? 3Fe^2+(aq) rightarrow Fe(s) + 2Fe^3+(aq) If not write the equation for the spontaneous reaction, calculate Edegree_Cell, and rank the three species of iron in order of decreasing reducing strength.

Explanation / Answer

1) Balancing the redox reaction in basic medium.

Given skeleton ionic reaction,

MnO4– (aq.) + I– (aq.) ------------> MnO42– (aq.) + IO3– (aq.)

Mn(+7)          I (–1)                       Mn(+6)             I (+5)

Two Half-cell reactions are,

Oxidation Half reaction (OHR): I– (aq.) ------------> IO3– (aq.)

Reduction Half reaction (RHR): MnO4– (aq.) ------------> MnO42– (aq.)

Let us balance these two half-cell reaction separately using following steps.

A) Balancing of atoms other than O and H.

No need as they are already in both OHR and RHR.

B) Balancing of O and H by adding H2O on O deficient side and H+ on other side:

OHR: we need to add to 3 H2O on left and 6 H+ on right side

OHR: I– (aq.) + 3 H2O ------------> IO3– (aq.) + 6 H+ (aq.)

RHR: No need as they are balanced already.

C) Balancing of charge by adding appropriate number of e– on appropriate side.

OHR: I– (aq.) + 3 H2O ------------> IO3– (aq.) + 6 H+ (aq.) + 5e–

RHR: MnO4– (aq.) + e– -----------> MnO42– (aq.)

D) Balancing of e– number.

OHR: I– (aq.) + 3 H2O ------------> IO3– (aq.) + 6 H+ (aq.) + 6e–

We need to multiply RHR by 6.

RHR: 6 MnO4– (aq.) + 6e– -----------> 6 MnO42– (aq.)

E) Let us add two balance half-cell reactions.

I– (aq.) + 3 H2O + 6 MnO4– (aq.) ------------> IO3– (aq.) + 6 H+ (aq.) + 6 MnO42– (aq.)

Electrons cancelled out mutually.

This is the balanced equation in acidic medium.

For the balanced equation in basic medium we need to add 6 OH– on right side to neutralize 6H+ ions. And for balance sake we need to add OH– on left as well.

6 H+ and 6 OH– ions forms 6 H2O on right 3 of which cancel 3 H2O on left. Finally,

I– (aq.) + 6 MnO4– (aq.) + 6 OH– (aq.) ------------> IO3– (aq.) + 6MnO42– (aq.) + 3H2O

Hence the required balanced equation in alkaline medium.

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