2) List the following ionic species as electron donor in descending order. I-Ba^+2(E^0Ba^2+/Ba_((s))=-2.906V) II-Cd^+2(E^0Cd^2+/Cd_((s))=-0.403V) III-Co^+3(E^0Co^3+/Co^2+=1.808V) IV-Cu^+2(E^0Cu^2+/Cu_((s))=0.337V) A) Igt IIgt IIgt gt VV B) IIgt Igt IVgt III C Igt IIgt IVgt III D) IIIgt IVgt IIgt I E) IIIgt IVgt Igt II

Question

2) List the following ionic species as electron donor in descending order. I-Ba^+2(E^0Ba^2+/Ba_((s))=-2.906V) II-Cd^+2(E^0Cd^2+/Cd_((s))=-0.403V) III-Co^+3(E^0Co^3+/Co^2+=1.808V) IV-Cu^+2(E^0Cu^2+/Cu_((s))=0.337V) A) Igt IIgt IIgt gt VV B) IIgt Igt IVgt III C Igt IIgt IVgt III D) IIIgt IVgt IIgt I E) IIIgt IVgt Igt II

Answer 1

To determine the order of ionic species as electron donors, we need to consider their standard reduction potentials (E° values). The higher the reduction potential, the greater the tendency of the species to gain electrons and act as an electron acceptor. Conversely, the lower the reduction potential, the greater the tendency of the species to lose electrons and act as an electron donor.Given the E° values:I. Ba²⁺: -2.906 VII. Cd²⁺: -0.403 VIII. Co³⁺: +1.808 VIV. Cu²⁺: +0.337 VThe order of ionic species as electron donors in descending order is:III (Co³⁺) > IV (Cu²⁺) > II (Cd²⁺) > I (Ba²⁺)Therefore, the correct answer is:D) III > IV > II > I

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