To determine the overall voltage for the redox reaction involving the half-reactions ( \text{Ag}^+ + e^- \rightarrow \text{Ag}(s) ) and ( \text{Cu}(s) \rightarrow \text{Cu}^{2+} + 2e^- ), we first need the standard reduction potentials. The standard reduction potential for silver (( \text{Ag}^+ )) is +0.80 V, and for copper (( \text{Cu}^{2+} )) is +0.34 V. Since silver is reduced and copper is oxidized, the overall cell potential is calculated as ( E_{\text{cell}} = E_{\text{reduction}} - E_{\text{oxidation}} = 0.80 , \text{V} - 0.34 , \text{V} = 0.46 , \text{V} ). Thus, the overall voltage for the redox reaction is +0.46 V.
0.34-(-2.37)
0.34-(-2.37)
0.34-(-2.37)
0.34 - (-2.37) you're welcome.
0.80-0.34
The overall voltage for the non-spontaneous redox reaction between Mg and Cu can be calculated by finding the difference in standard reduction potentials between the two half-reactions. The standard reduction potentials for Mg and Cu are -2.37 V and 0.34 V, respectively. Therefore, the overall voltage would be (-2.37 V) - (0.34 V) = -2.71 V.
-2.37 - 0.34
The overall redox reaction of Cr2O7 + Br is not a balanced equation. To balance the equation, the half-reactions for the oxidation and reduction of each element need to be determined and balanced first.
The overall voltage for the redox reaction involving Ag and Cu is determined by subtracting the reduction potential of the anode from the reduction potential of the cathode. Given the reduction potentials of Ag and Cu as 0.80 V and 0.34 V respectively, the overall voltage is 0.46 V, calculated as (0.34 V) - (0.80 V).
No, single displacement and double displacement reactions are not always redox reactions. Redox reactions involve electron transfer between reactants, while single displacement and double displacement reactions do not always involve the transfer of electrons.
Single displacement reaction and a double displacement reaction are redox reactions. apex- false
0.34 - (-2.37) you're welcome.