Q - An electrochemical reaction occurs between an unknown element and zinc. The half-cell reaction for the zinc is:
Zn(s) → Zn2+(aq) + 2e−
The cell potential for the reaction is Eºcell = 1.83 V.
Is the reaction spontaneous or nonspontaneous? Explain.
What is the half-cell potential of the unknown species?
What is the identity of the unknown element?
I think that the reaction is spontaneous because no outside force is required for it to take place. The half-cell potential is 0.915 and the unknown element is Mercury. Could you check my answers and help if I got them wrong?
To determine if a reaction is spontaneous or nonspontaneous, we can use the cell potential (Eºcell). If the cell potential is positive, then the reaction is spontaneous, while a negative cell potential indicates a nonspontaneous reaction.
In this case, the cell potential (Eºcell) is given as 1.83 V, which is a positive value. Therefore, the reaction is indeed spontaneous, as you correctly pointed out.
The half-cell potential of the unknown species can be determined by using the Nernst equation, which relates the cell potential to the concentrations of reactants and products involved in the half-cell reactions. The Nernst equation can be written as:
E = Eº - (RT/nF) * ln(Q)
Where:
E = cell potential
Eº = standard cell potential
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin
n = number of moles of electrons transferred
F = Faraday's constant (96,485 C/mol)
ln = natural logarithm
Q = reaction quotient (ratio of product concentrations to reactant concentrations)
Since we are considering the half-cell reaction given by:
Zn(s) → Zn2+(aq) + 2e−
The reaction quotient (Q) would be [Zn2+(aq)]/ [Zn(s)], and since the solid form of zinc is in its standard state, the concentration of Zn(s) is typically taken as 1M, so Q becomes [Zn2+(aq)]/1.
Since the Nernst equation relates the cell potential (E) to the logarithm of Q, and we know the values of Eºcell (1.83 V) and Q (which is equal to the concentration of Zn2+(aq)), we can rearrange the Nernst equation to solve for E:
E = Eº - (RT/nF) * ln(Q)
By substituting the given values into the equation, we can solve for E:
E = 1.83 V - ((8.314 J/mol·K)(298 K) / (2 mol electrons)(96,485 C/mol)) * ln([Zn2+(aq)])
Once we know the value of E, we can compare it to the standard reduction potentials (Eº) table to determine the identity of the unknown element. The element with a matching potential will be the unknown element.
Since I don't have the exact concentration of Zn2+(aq) or the temperature, I cannot calculate the exact value of the half-cell potential or identify the unknown element. It's advisable to use the given concentration and temperature to complete the calculations and refer to the standard reduction potentials table to identify the element.