Question

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    

(14分)某中學(xué)課外興趣小組用惰性電極電解飽和食鹽水（含少量Ca²⁺、Mg²⁺)作系列探究，裝置如下圖所示：

(1)電解時，甲同學(xué)發(fā)現(xiàn)電極a附近溶液出現(xiàn)渾濁,請用離子方程式表示原因____________

(2)一段時間后，你認(rèn)為C中溶液可能出現(xiàn)的現(xiàn)象是______，請用離子方程式表示原因________________________

(3) 實(shí)驗(yàn)結(jié)束后，乙同學(xué)將A中的物質(zhì)冷卻后加入到H₂S溶液中發(fā)現(xiàn)有氣泡出現(xiàn)，但加人到稀鹽酸中卻沒有任何現(xiàn)象。請用化學(xué)方程式和簡要的文字解釋原因____________

(4) 隨著反應(yīng)的進(jìn)行，興趣小組的同學(xué)們都特別注意到D中溶液紅色逐漸褪去。他們對溶液紅色褪去主要原因提出了如下假設(shè)，請你完成假設(shè)二。

假設(shè)一:B中逸出氣體與水反應(yīng)生成的物質(zhì)有強(qiáng)氧化性，使紅色逐漸褪去；

假設(shè)二 ：__________________________________________。

(5) 請你設(shè)計(jì)實(shí)驗(yàn)驗(yàn)證上述假設(shè)一，寫出實(shí)驗(yàn)步驟及結(jié)論___________________________________________________________________________

 

Answer 1

【答案】

（共 14 分）

（1）2H⁺ + 2e^-=H₂↑,Mg²⁺ + 2OH^-=Mg（OH）₂↓（2分）

（2）有淡黃色沉淀（2分）   Cl₂+S^2-=2Cl^-+S↓（2分）

（3）Ag置換H₂S的反應(yīng)式為：2Ag+H₂S=Ag₂S+H₂,表明Ag₂S的溶解度比AgCl小很多（3分）

（4）可能是H⁺中和了OH^-造成褪色（2分）

（5）1.取少量D中褪色后的溶液于試管中

2.滴加1-2滴酚酞溶液,如果又顯紅色,則假設(shè)一正確（3分）

【解析】略