Write net equations for the spontaneous redox reactions that occur during the fo

Question

Write net equations for the spontaneous redox reactions that occur during the following or none if there is no extensive reaction. Instructions: Enter all substances in the order listed at the top of the column. Use a carot to indicate a superscript, but do nothing for subscripts. Use a hyphen + greater than (->) for yields. Click on the eye symbol to check your formatting.

Reactants Rxn: Ox(1) + Red(2) + other -> Red(1) + Ox(2) + other Iron metal is dipped into a NiSO4 solution Chromium metal is placed in hydrochloric acid A copper wire is added to nitric acid
Magnesium metal is added to water
A drop of mercury falls into an acidified solution of KMnO4

Explanation / Answer

Solution:- (1) standard reduction potential for Fe is greater than Ni. So, when Iron metal is dipped into NiSO4 solution then reduction of Ni and oxidation of Fe takes place and the equation could be written as...

Ni2+(aq) + Fe(s) -> Ni(s) + Fe2+(aq)

(2) Chromium metal is placed above hydrogen in electrochemical series that means it has the tendency to replace hydrogen gas from hydrogen acids. The net redox equation could be written as...

6H+(aq) + 2Cr(s) -> 3H2(g) + 2Cr3+(aq)

(3) in electrochemical series, Copper is placed below hydrogen that means it can not replace hydrogen from hydrogen acids and so there will be no reaction when Copper wire is added to nitric acid.

H+(aq) + Cu(s) -> none

(4) Magnesium is highly reactive and when it is added to water then hydrogen gas is released from water.

2H2O(l) + Mg(s) -> H2(g) + Mg2+(aq) + 2OH-(aq)

(5) Hg metal is placed above Mn so Mg will oxidized and Mn present in KMnO4 is reduced.

KMnO4 breaks into K+ and MnO4-. In MnO4-, Mn is present in +7 oxidation state. It's half equation could be written as..

8H+(aq) + MnO4-(aq) + 5e- -> Mn2+(aq) + 4H2O(l)

and the half equation for the oxidation of Hg could be written as..

Hg(l) -> Hg2+(aq) + 2e-

two make the electrons equal we need to multiply first half equation by 2 and second half equation by 5

16H+(aq) + 2MnO4-(aq) + 10e- -> 2Mn2+(aq) + 8H2O(l)

5Hg(l) -> 5Hg2+(aq) + 10e-

on adding these two hale equations...

2MnO4-(aq) + 5Hg(l) + 16H+(aq) -> 2Mn2+(aq) + 5Hg2+(aq) + 8H2O(l)

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