cr(s)--->Cr^3(aq)+3e- and Pb2+(aq)+2e---->Pb(s) write a balanced equation
Please help thanks
To write a balanced equation, we need to determine the stoichiometric coefficients for each reactant and product.
For the first half-reaction, we have:
Cr(s) --> Cr^3+(aq) + 3e-
Since the chromium (Cr) loses 3 electrons, we need to balance the equation by multiplying Cr on the left side by 3:
3Cr(s) --> 3Cr^3+(aq) + 9e-
For the second half-reaction, we have:
Pb^2+(aq) + 2e- --> Pb(s)
The equation is already balanced for the number of electrons.
Now, we can combine the two half-reactions and balance the overall equation:
3Cr(s) + 2Pb^2+(aq) + 6e- --> 3Cr^3+(aq) + 2Pb(s) + 6e-
By canceling out the number of electrons on both sides:
3Cr(s) + 2Pb^2+(aq) --> 3Cr^3+(aq) + 2Pb(s)
This is the balanced equation for the reaction.