Given these standard reduction potentials, what is the standard reduction potential for
`Co^(3+)(aq) + 3e^(-) rightarrow Co(s)`?
`Co^(3+)(aq) + e^(-) rightarrow Co^(2+)(aq) E^(@) = 1.82V`
`Co^(2+)(aq) + 2e^(-) rightarrow Co(s) E^(@) = -0.28V`

Given the standard reduction potentials, which statement is correct? {:(Cu^(2)(aq) + 2e^(-) rightarrow Cu(s), E^(@) = 0.34V), (2H^(+)(aq) + 2e^(-) rightarrow H_(2)(g), E^(@)= 0.0V), (Cr^(3+)(aq) + 2e^(-) rightarrow Cr(s),E^(@) = -0.73V):}

Given these standards reduction potentials, what is the free energy change (in kJ. Mol^(-) for the reaction: Pb(s) + 2Ag^(+)(aq) rightarrow Pb^(2+)(aq) + 2Ag(s) Ag^(+)(aq) + e^(-) rightarrow Ag(s) E^(@) = 0.80V Pb^(2+)(aq) + 2e^(-) rightarrow Pb(s) E^(@) = -0.13V

Given the two standard reduction potentials below, what is the K_(sp) of Ag_(2)CrO_(4) at 25^(@)C ? Ag_(2)CrO^(4)(s) + 2e^(-) rightarrow 2Ag(s) + CrO_(4)^(2-)(aq) E^(@)= + 0.446V Ag^(+)(aq) + e^(-) rightarrow Ag(s) E^(@) = +0.799V

According to the standard reduction potentials: Pb^(2+)(aq) + 2e^(-) rightarrow Pb(s) E^(@) = -0.13V Fe^(2+)(aq) + 2e^(-) rightarrow Fe(s) E^(@) = -0.44 Zn^(2+)(aq) + 2e^(-) rightarrow Zn(s) E^(@) = -0.76V Which species will reduce Mn^(3+) to Mn^(2+) [E^(@) = 1.51V but will NOT reduce Cr^(3+) to Cr^(2+) [E^(@) = -0.40V]

Use the given standard reduction potentials to determine the reduction potential for this half-reactions. MnO_(4)^(-)(aq) + 3e^(-) + 4H^(+) rightarrow MnO_(2)(S) + 2H_(2)O(l)

Which of the half reactions, when coupled, will make a galvanic cell that will produce the largest voltage under standard conditons? P. Cu^(2+)(aq) + 2e^(-) rightarrow Cu(s) E^(@) = +0.34V Q. Pb^(2+)(aq) + 2e^(-) rightarrow Pb(s) E^(@) = -0.13V R. Ag^(+)(aq) + 3e^(-) rightarrow Al(s) E^(@) = -1.66V

What is the standard cell potential for the reaction, 2Cr(s) + 3Sn^(2+)(aq) rightarrow 3Sn(s) + 2Cr^(3+)(aq) given the E^(@) values shown?

Use the standard reduction potentials: Sn^(2+)(aq) + 2e^(-) rightarrow Sn(s) E^(@) = -0.141V Ag^(+)(aq) + e^(-) rightarrow Ag(s) E^(@) = 0.800V To calcultate E^(@) for the reaction: Sn(s) + 2Ag^(+)(aq) rightarrow Sn^(2+)(aq) + 2Ag(s)

Use the standard reduction potentials to determine what is observed at the cathode during the electrolysis of a 1.0 M solution of KBr that contains phenolphthalein. O_(2)(g) + 4H^(+)(aq) + 4e^(-) rightarrow 2H_(2)O(l) E^(@) = 1.23V Br_(2)(l) + 2e^(-) rightarrow 2Br^(-)(aq) E^(@) = 1.07V 2H_(2)O(l) + 2e^(-) rightarrow H_(2)(g) + 2OH^(-) E^(@0 = -0.80V K^(+) (aq) + e^(-) rightarrow K(s) E^(@) = -2.92V