cr(s)--->Cr^3(aq)+3e- and Pb2+(aq)+2e---->Pb(s) write a balanced equation
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To balance the given half-reactions:
1. Start by balancing all the elements except oxygen and hydrogen.
cr(s) → Cr^3+(aq) + 3e-
Pb^2+(aq) + 2e- → Pb(s)
2. Balance the oxygen atoms by adding water molecules (H2O) as needed:
cr(s) → Cr^3+(aq) + 3e-
Pb^2+(aq) + 2e- → Pb(s) + 2H2O(l)
3. Balance the hydrogen atoms by adding hydrogen ions (H+) as needed:
cr(s) → Cr^3+(aq) + 3e-
Pb^2+(aq) + 2e- → Pb(s) + 2H2O(l) + 2H+(aq)
4. Balance the charges by multiplying the half-reactions by appropriate coefficients:
3(cr(s) → Cr^3+(aq) + 3e-)
2(Pb^2+(aq) + 2e- → Pb(s) + 2H2O(l) + 2H+(aq))
Finally, the balanced equation is:
3cr(s) + 2Pb^2+(aq) + 6H+(aq) → 3Cr^3+(aq) + 2Pb(s) + 3H2O(l)
To balance the given redox reactions, you can follow the steps below:
1. Identify the oxidation states (or oxidation numbers) of each element in the reaction.
In the first equation:
Cr in Cr(s) has an oxidation state of 0.
Cr in Cr^3+(aq) has an oxidation state of +3.
In the second equation:
Pb in Pb2+(aq) has an oxidation state of +2.
Pb in Pb(s) has an oxidation state of 0.
2. Write the half-reactions for the oxidation and reduction processes.
Oxidation half-reaction: Cr(s) --> Cr^3+(aq) + 3e-
Reduction half-reaction: Pb2+(aq) + 2e- --> Pb(s)
3. Balance the number of electrons in each half-reaction.
In the oxidation half-reaction, there are 3 electrons on the right side (3e-), so multiply the reduction half-reaction by 3 to balance the number of electrons.
Oxidation half-reaction: Cr(s) --> Cr^3+(aq) + 3e-
Reduction half-reaction (multiplied by 3): 3Pb2+(aq) + 6e- --> 3Pb(s)
4. Balance the atoms other than hydrogen and oxygen.
In the reduction half-reaction, there are now 3 Pb atoms on the right side. To balance the number of Pb atoms in the oxidation half-reaction, include the same number on the left side.
Balanced oxidation half-reaction: 3Cr(s) --> 3Cr^3+(aq) + 9e-
Balanced reduction half-reaction: 3Pb2+(aq) + 6e- --> 3Pb(s)
5. Finally, add the two balanced half-reactions together.
3Cr(s) + 3Pb2+(aq) + 9e- --> 3Cr^3+(aq) + 3Pb(s) + 6e-
Cancel out the electrons on both sides of the equation:
3Cr(s) + 3Pb2+(aq) --> 3Cr^3+(aq) + 3Pb(s)
This is the balanced equation for the given redox reactions.