Thank you for the help. There are two questions. The first I have answered, but

Question

Thank you for the help. There are two questions. The first I have answered, but it is needed for the second question.

Question 1.

An electrochemical cell is created using gold and magnesium half-cells under standard conditions (ie. A gold bar in a gold ion solution, and a magnesium strip in a magnesium ion solution).

Write the equations for the half-cell undergoing oxidation and undergoing reduction

Write the overall, balanced equation

Identify the anode and cathode

Calculate the voltage for the cell

You do not need to diagram the cell

Answer to Question 1.

Question 2.

Consider the same electrochemical cell as in the question above.

Imagine that after the cell has run for some time, the mass of the magnesium electrode has changed by 5.0 g. What must have been the change in mass of the gold electrode, and will its mass increase or decrease?

Attempt at answer:

"As the electrons will be deposited at the cathode, Mg2+ will gain 5g and Au (the gold electrode) will lose 5g."

But I need to include stoichiometry in my answer and I'm not sure how. Thank you for any help!

Explanation / Answer

the complete cell reaction is

2 Au+3 + 3Mg(s) ----> 2Au + 3 Mg+2

from the above reaction

moles of Au+3 reacted = (2/3) x moles of Mg

given

5 g of Mg(s) is oxidized

we know that

moles = mass / molar mass

also

molar mass of Mg is 24 g/mol

so

moles of Mg = 5 / 24

moles of Mg = 0.20833333

moles of Au+3 reacted = (2/3) x moles of Mg

moles of Au+3 reacted = (2/3) x 0.208333

moles of Au+3 reacted = 0.138888

now

mass of Au+3 reacted = moles x molar mass

mass of Au+3 reacted= 0.13888 x 197

mass of Au+3 reacted= 27.36 g

so

there is a change of 27.36 g in the mass of the gold electrode

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