Q: For the redox reaction, \$MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O\$, the correct coefficients of the reactants for the balanced equation are _____________.

The unbalanced redox reaction is \$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O\$Balance C atoms.\$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + 2CO_{2} + H_{2}O\$The oxidation number of Mn decreases from \$\displaystyle +7\$ to \$\displaystyle +2\$.The decrease in the oxidation number is \$\displaystyle 7-2=5\$The oxidation number of C increases from \$\displaystyle +3\$ to \$\displaystyle +4\$.The increase in the oxidation number for 1 C atom is \$\displaystyle 4-3=1\$The increase in the oxidation number for 2 C atom is \$\displaystyle 2 \times 1=2\$To balance the increase in the oxidation number with decrease in the oxidation number, multiply Mn containing species with 2 and C containing species with 5.\$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} \$ Balance O atoms by adding 8 water molecules to the products side.\$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O\$ Balance H atoms by adding 16 \$\displaystyle H^+\$ ions to the reactants side.\$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+}\rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O\$ This is the balanced equation.

Q: \$NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + H_{2}O\$. Balance the given chemical equation

The balanced chemical equation is as follow:\$2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O\$.

Q: The oxidation number of Ni in \$Ni(CO)_4\$ is \$+2\$.

Let \$x\$ be the oxidation number of \$Ni\$ in \$Ni(CO)_4\$.Since the overallcharge on the complex is \$0\$, the sum of oxidation states of all elements init should be equal to \$0\$.Therefore, \$x + 4 (0) = 0\$Hence, \$x = 0\$.Thus, the oxidation number of \$Ni\$ in \$Ni(CO)_4\$ is \$0\$.Hence, the correct option is \$\text{B}\$

Q: Balance the following redox reactions by the ion-electron method in acidic medium.\$MnO^{-}_4(aq)+SO_2(g)\rightarrow Mn^{2+}(aq)+HSO^{-}_4(aq)\$

The unbalanced chemical equation is:\$\displaystyle MnO^{-}_4(aq)+SO_2(g)\rightarrow Mn^{2+}(aq)+HSO^{-}_4(aq) \$The oxidation half reaction is \$\displaystyle SO_2(g) +2H_2O(l) \rightarrow HSO_4^-(aq) +3H^+(aq) +2e^-(aq)\$.The reduction half reaction is \$\displaystyle MnO_4^-(aq) \rightarrowMn^(2+)(aq) \$.In the reduction half reaction, the oxidation number of Mn changes from +7 to +2. Hence, 5 electrons are added to LHS of the reaction.\$\displaystyle MnO_4^-(aq) +5e^- \rightarrow Mn^{2+}(aq) \$Charge is balanced in the reduction half reaction by adding 8 hydrogen ions to LHS.\$\displaystyle MnO_4^-(aq) +5e^- +8H^+(aq) \rightarrow Mn^{2+}(aq) \$To balance O atoms, 4 water molecules are added on RHS.\$\displaystyle MnO_4^-(aq) +5e^- +8H^+(aq) \rightarrow Mn^{2+}(aq) +4H_2O(l) \$To equalize the number of electrons, the oxidation half reaction is multiplied by 5 and the reduction half reaction is multiplied by 2.\$\displaystyle 5SO_2(g) +10H_2O(l) \rightarrow 5HSO_4^-(aq) +15H^+(aq) +10e^-(aq)\$\$\displaystyle 2MnO_4^-(aq) +10e^- +16H^+(aq) \rightarrow 2Mn^{2+}(aq) \$Two half cell reactions are added to obtain a balanced equation.\$\displaystyle 2MnO^{-}_4(aq)+5SO_2(g)+2H_2O(l)\rightarrow 2Mn^{2+}(aq)+5HSO^{-}_4(aq) \$

Q: Balance the following redox reactions by ion electron method.\$Cr_2O^{2-}_7+SO_2(g)\rightarrow Cr^{3+}(aq)+SO^{2-}_4(aq)\$ (in acidic solution)

The oxidation half reaction is \$\displaystyle SO_2(g) + 2H_2O(l) \rightarrow SO_4^{2-} + 4H^+(aq) +2e^- \$.The reduction half reaction is \$\displaystyle Cr_2O_7^{2-}(aq) + 14H^+(aq) +6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l) \$.The oxidation half reaction is multiplied by 3 and added to the reduction half reaction to obtain the balanced redox reaction.\$\displaystyle Cr_2O^{2-}_7+3SO_2(g)+2H^+(aq) \rightarrow 2Cr^{3+}(aq)+3SO^{2-}_4(aq)+H_2O(l) \$

Q: Permanganate ion reacts with bromide ion in basic medium to give magnesium dioxide and bromate ion.Write the balanced ionic equation for the reaction.

1)\$2MnO_4{^-}(aq)+Br-(aq) +H_2O(l) \longrightarrow 2MnO_2(s)+BrO_3(aq) + 2OH^-(aq)\$Here, instead of solving the overall reaction we solve as half reactions. The two half reactions occurring here are written as:\$Br^- + 3H_2O \longrightarrow BrO_3{^-} + 6H^+ + 6e^-\$ -----(Oxidation)\$MnO_4{^-} + 4H^+ + 3e^- → MnO_2 +2H_2O\$----- (Reduction)As in the reduction reaction no. of e- involved is half that of the one in the oxidation we will multiply the reaction by 2 so it becomes \$2MnO_4{^-} + 8H^+ + 6e^- \longrightarrow 2MnO_2 + 4H_2O\$Thus, reaction can be balanced.

Question 1

For the redox reaction, $MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O$,  the correct coefficients of the reactants for the balanced equation are _____________.

Accepted Answer

The unbalanced redox reaction is $\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O$Balance C atoms.$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + 2CO_{2} + H_{2}O$The oxidation number of Mn decreases from $\displaystyle +7$ to $\displaystyle +2$.The decrease in the oxidation number is $\displaystyle 7-2=5$The oxidation number of C increases from $\displaystyle +3$ to $\displaystyle +4$.The increase in the oxidation number for 1 C atom  is $\displaystyle 4-3=1$The increase in the oxidation number for 2 C atom  is $\displaystyle 2 \times 1=2$To balance the increase in the oxidation number with decrease in the oxidation number, multiply Mn containing species with 2 and C containing species with 5.$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} $ Balance O atoms by adding 8 water molecules to the products side.$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$ Balance H atoms by adding 16 $\displaystyle H^+$ ions to the reactants side.$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+}\rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$ This is the balanced equation.

Question 2

$NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + H_{2}O$. Balance the given chemical equation

Accepted Answer

The balanced chemical equation is as follow:$2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O$.

Question 3

In the reaction $ \rightarrow $$ MNO_2 +4 HCL \rightarrow MnCl_2 +2 H_2O +Cl_2 $(a) name the compound (i)  oxidised  (ii) reduced (b) define oxidation of reduction on its basis.

Accepted Answer

Here is your answer -----a. Here HCl is oxidized to Cl2 and Mno2 is reduced to MnCl2.b. The reaction in which oxygen is either gained or hydrogen is lost by a substance is called oxidation reaction.The reaction in which hydrogen is gained or oxygen is lost by a substance is called reduction reaction.

Question 4

The oxidation number of Ni in $Ni(CO)_4$ is $+2$.

Accepted Answer

Let $x$ be the oxidation number of $Ni$ in $Ni(CO)_4$.Since the overallcharge on the complex is $0$, the sum of oxidation states of all elements init should be equal to $0$.Therefore, $x + 4 (0) = 0$Hence, $x = 0$.Thus, the oxidation number of $Ni$ in $Ni(CO)_4$ is $0$.Hence, the correct option is $\text{B}$

Question 5

Balance the following redox reactions by the ion-electron method in acidic medium.$MnO^{-}_4(aq)+SO_2(g)\rightarrow Mn^{2+}(aq)+HSO^{-}_4(aq)$

Accepted Answer

The unbalanced chemical equation is:$\displaystyle MnO^{-}_4(aq)+SO_2(g)\rightarrow Mn^{2+}(aq)+HSO^{-}_4(aq) $The oxidation half reaction is $\displaystyle SO_2(g) +2H_2O(l) \rightarrow HSO_4^-(aq) +3H^+(aq) +2e^-(aq)$.The reduction half reaction is $\displaystyle MnO_4^-(aq) \rightarrowMn^(2+)(aq) $.In the reduction half reaction, the oxidation number of Mn changes from +7 to +2. Hence, 5 electrons are added to LHS of the reaction.$\displaystyle MnO_4^-(aq) +5e^- \rightarrow Mn^{2+}(aq) $Charge is balanced in the reduction half reaction by adding 8 hydrogen ions to LHS.$\displaystyle MnO_4^-(aq) +5e^- +8H^+(aq) \rightarrow Mn^{2+}(aq) $To balance O atoms, 4 water molecules are added on RHS.$\displaystyle MnO_4^-(aq) +5e^- +8H^+(aq) \rightarrow Mn^{2+}(aq) +4H_2O(l) $To equalize the number of electrons, the oxidation half reaction is multiplied by 5 and the reduction half reaction is multiplied by 2.$\displaystyle 5SO_2(g) +10H_2O(l) \rightarrow 5HSO_4^-(aq) +15H^+(aq) +10e^-(aq)$$\displaystyle 2MnO_4^-(aq) +10e^- +16H^+(aq) \rightarrow 2Mn^{2+}(aq) $Two half cell reactions are added to obtain a balanced equation.$\displaystyle 2MnO^{-}_4(aq)+5SO_2(g)+2H_2O(l)\rightarrow 2Mn^{2+}(aq)+5HSO^{-}_4(aq) $

Question 6

Balance the following redox reactions by ion electron method.$Cr_2O^{2-}_7+SO_2(g)\rightarrow Cr^{3+}(aq)+SO^{2-}_4(aq)$ (in acidic solution)

Accepted Answer

The oxidation half reaction is $\displaystyle SO_2(g) + 2H_2O(l) \rightarrow SO_4^{2-} + 4H^+(aq) +2e^- $.The reduction half reaction is $\displaystyle Cr_2O_7^{2-}(aq) + 14H^+(aq) +6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l) $.The oxidation half reaction is multiplied by 3 and added to the reduction half reaction to obtain the balanced redox reaction.$\displaystyle Cr_2O^{2-}_7+3SO_2(g)+2H^+(aq) \rightarrow 2Cr^{3+}(aq)+3SO^{2-}_4(aq)+H_2O(l) $

Question 7

In the reaction $CuO + H_2 \rightarrow Cu + H_2O$ which substance gets oxidised and which gets reduced?Give one more example of such type of reaction.

Accepted Answer

An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. In redox reaction the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.In the reaction : ${ CuO\quad +\quad { H }_{ 2 } }\quad \longrightarrow \quad Cu\quad +\quad { H }_{ 2 }O$ n the reaction ${ H }_{ 2 } $  is oxidized since it has lost electrons and $Cu$ is reduced since it has gained electrons.

Question 8

What is redox reaction? Explain with one example.

Accepted Answer

An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.The formation of hydrogen fluoride is an example of a redox reaction. We can break the reaction down to analyze the oxidation and reduction of reactants. The hydrogen is oxidized and loses two electrons, so each hydrogen becomes positive.$H_2+2F \rightarrow 2HF$

Question 9

Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent. $P_4(s)+OH^{-}(aq)\rightarrow PH_3(g)+H_2PO^{-}_2(aq)$

Accepted Answer

Oxidation number method: The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, $\displaystyle P_4 $ is oxidizing as well as reducing agent.During reduction, the total decrease in the oxidation number for 4 P atoms is 12.During oxidation, total increase in the oxidation number for 4 P atoms is 4.The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $\displaystyle H_2PO_2^-$ with 3. $\displaystyle P_4(s) +OH^-(aq) \rightarrow PH_3(g) + 3 H_2PO_2^-(aq) $To balance O atoms, multiply $\displaystyle OH^- $ ions by 6.$\displaystyle P_4(s) +6OH^-(aq) \rightarrow PH_3(g) + 3 H_2PO_2^-(aq) $To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.$\displaystyle P_4(s) +6OH^-(aq) +3H_2O(l)\rightarrow PH_3(g) + 3 H_2PO_2^-(aq) +3OH^-(aq)$Subtract 3 hydroxide ions from both sides.$\displaystyle P_4(s) +3OH^-(aq) +3H_2O(l)\rightarrow PH_3(g) + 3 H_2PO_2^-(aq) $Ion electron method:The oxidation half reaction is $\displaystyle P_4(s) \rightarrow H_2PO_2^-(aq)  $.The P atom is balanced.$\displaystyle P_4(s) \rightarrow 4H_2PO_2^-(aq)  $The oxidation number is balanced by adding 4 electrons on RHS.$\displaystyle P_4(s) \rightarrow 4H_2PO_2^-(aq)  +4e^-$The charge is balanced by adding 8 hydroxide ions on LHS.$\displaystyle P_4(s) +8OH^-(aq)\rightarrow 4H_2PO_2^-(aq)  $The O and H atoms are balanced.The reduction half reaction is $\displaystyle P_4(s) \rightarrow PH_3(g) $.The oxidation number is balanced by adding 12 eelctrons on LHS.$\displaystyle P_4(s) +12e^- \rightarrow PH_3(g) $The charge is balanced by adding 12 hydroxide ions on RHS.$\displaystyle P_4(s) +12e^- \rightarrow PH_3(g)+12OH^- $The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2. The half reactions are then added to obtain balanced chemical equation.$\displaystyle P_4(s) + 3OH^-(aq) +3H_2O (l) \rightarrow PH_3(g)+3H_2PO_2^- (aq) $

Question 10

Permanganate ion reacts with bromide ion in basic medium to give magnesium dioxide and bromate ion.Write the balanced ionic equation for the reaction.

Accepted Answer

1)$2MnO_4{^-}(aq)+Br-(aq) +H_2O(l) \longrightarrow 2MnO_2(s)+BrO_3(aq) + 2OH^-(aq)$Here, instead of solving the overall reaction we solve as half reactions. The two half reactions occurring here are written as:$Br^- + 3H_2O \longrightarrow BrO_3{^-} + 6H^+ + 6e^-$ -----(Oxidation)$MnO_4{^-} + 4H^+  + 3e^- &#8594; MnO_2 +2H_2O$----- (Reduction)As in the reduction reaction no. of e- involved is half that of the one in the oxidation we will multiply the reaction by 2 so it becomes $2MnO_4{^-} + 8H^+ + 6e^- \longrightarrow 2MnO_2 + 4H_2O$Thus, reaction can be balanced.