This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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bazuunelson, let's knock this out. Here's the solution to question 11: 11. Study the electrode potentials in the table below and answer the question that follow: H^2+(aq) + 2e^- H(s) E^ = +0.34 V Z^2+(aq) + 2e^- Z(s) E^ = -2.38 V G^+(aq) + e^- G(s) E^ = +0.80 V T^2+(aq) + 2e^- T(s) E^ = -2.87 V a) Which one is the strongest reducing agent? Step 1: Understand reducing agents. A reducing agent is a substance that readily loses electrons (gets oxidized). The stronger the reducing agent, the more negative its standard reduction potential (E^). Step 2: Compare the E^ values for the elemental forms (H, Z, G, T). The most negative E^ value corresponds to the strongest reducing agent. H: +0.34 V Z: -2.38 V G: +0.80 V T: -2.87 V Step 3: Identify the strongest reducing agent. The most negative E^ value is -2.87 V, which corresponds to T. a) The strongest reducing agent is: T b) Write the ionic equation for the reaction that takes place when Z is dipped in a solution of G^+ ions. Step 1: Determine which species will be oxidized and which will be reduced. Compare the E^ values for Z^2+/Z and G^+/G. Z^2+(aq) + 2e^- Z(s) E^ = -2.38 V G^+(aq) + e^- G(s) E^ = +0.80 V Since Z has a more negative E^ value than G, Z will be oxidized (lose electrons) and G^+ ions will be reduced (gain electrons). Step 2: Write the oxidation and reduction half-reactions. Oxidation (at anode): Z(s) Z^2+(aq) + 2e^- Reduction (at cathode): G^+(aq) + e^- G(s) Step 3: Balance the electrons and combine the half-reactions. Multiply the reduction half-reaction by 2 to balance the electrons: 2G^+(aq) + 2e^- 2G(s) Add the oxidation and balanced reduction half-reactions: Z(s) + 2G^+(aq) Z^2+(aq) + 2G(s) c) Calculate the E^ cell value of the reaction in 22.(b) above. (Assuming "22.(b)" refers to the reaction in 11.(b)) Step 1: Identify the standard electrode potentials for the oxidation and reduction half-reactions. From part (b), the reaction is: Z(s) + 2G^+(aq) Z^2+(aq) + 2G(s) Oxidation of Z: E^_anode = E^(Z^2+/Z) = -2.38 V Reduction of G^+: E^_cathode = E^(G^+/G) = +0.80 V Step 2: Calculate the standard cell potential (E^_cell). E^_cell = E^_cathode - E^_anode E^_cell = (+0.80 V) - (-2.38 V) E^_cell = +0.80 V + 2.38 V E^_cell = 3.18 V c) The E^ cell value is: 3.18 V Got more? Send 'em!