Lead metal is added to 0.140M Cr^+3(aq).
Pb(s)+ 2Cr^3+(aq)---> Pb^2+(aq) + 2Cr^2+(aq)
a)What is [Pb^2+] when equilibrium is established in the reaction?
b)What is [Cr^2+] when equilibrium is established in the reaction?
c)What is [Cr^3+] when equilibrium is established in the reaction?
Here is what I would do.
Look up standard potentials for Pb to Pb^+2 and for Cr^+3 to Cr+2.
Add the reduction half to the oxidation half to obtain Ecell at standard conditions.
At standard conditions,
nFEcell = 2.303*RT*log K
F is 96,485; R is 8.314 and T is 298. Calculate K, then set up an ICE chart for the reaction and solve for the components using the K value. Check my thinking.
To find the concentrations of the different species when equilibrium is established in the reaction, we can use the concept of stoichiometry and the expression for the equilibrium constant. Here's how you can approach each part of the question:
a) To find [Pb^2+], we need to use the stoichiometric relationship provided by the balanced equation. It states that 1 mole of Pb(s) reacts with 2 moles of Cr^3+. Assume x moles of Pb^2+ are formed at equilibrium. Since there is a 1:1 stoichiometry between Pb(s) and Pb^2+, the concentration of Pb^2+ will also be x M.
b) To find [Cr^2+], we need to use the stoichiometric relationship provided by the balanced equation. It states that 1 mole of Pb(s) reacts with 2 moles of Cr^3+, resulting in the formation of 2 moles of Cr^2+. So, the concentration of Cr^2+ will be 2x M.
c) To find [Cr^3+], we need to use the initial concentration of Cr^3+ and subtract the amount that reacts with the Pb(s). The initial concentration is given as 0.140 M. From the balanced equation, we know that the Pb(s) reacts in a 1:2 ratio with Cr^3+. Therefore, for every mole of Pb(s) that reacts, 2 moles of Cr^3+ are consumed. Since the moles of Pb(s) that react are the same as the moles of Cr^2+ formed, we can say that the concentration of Cr^3+ at equilibrium is [Cr^3+] = 0.140 M - (2 * x).
To find the value of x, we need to use the equilibrium expression for the reaction, which is given by K = [Pb^2+][Cr^2+]^2 / [Cr^3+]^2. Substituting the values we found in the previous steps, we can write the expression as K = (x)(2x)^2 / ([Cr^3+] - 2x)^2.
You can solve this equation for x by plugging in the numerical value of the equilibrium constant (K) and then use algebraic methods such as factoring or quadratic formula to solve for x. Once you have found the value of x, you can substitute it back into the equations in parts a, b, and c to obtain the final concentrations [Pb^2+], [Cr^2+], and [Cr^3+] at equilibrium.