1968 United States Senate election in Iowa

Summary

The 1968 United States Senate election in Iowa took place on November 5, 1968. Incumbent Republican U.S. Senator Bourke B. Hickenlooper retired. The open seat was won by Democratic Governor Harold E. Hughes, narrowly defeating Republican State Representative David M. Stanley.

1968 United States Senate election in Iowa

← 1962 November 5, 1968 1974 →
 
Nominee Harold E. Hughes David M. Stanley
Party Democratic Republican
Popular vote 574,884 568,469
Percentage 50.25% 49.69%

County results
Hughes:      50–60%      60–70%
Stanley:      50-60%      60-70%      70-80%

U.S. senator before election

Bourke B. Hickenlooper
Republican

Elected U.S. Senator

Harold E. Hughes
Democratic

General election edit

Candidates edit

Results edit

1968 United States Senate election in Iowa[1][2][3]
Party Candidate Votes % ±%
Democratic Harold E. Hughes 574,884 50.25%  3.64
Republican David M. Stanley 568,469 49.69%  3.70
Prohibition Verne Higens 727 0.06% N/A
Total votes 1,143,353 100.00%
Democratic gain from Republican Swing

See also edit

References edit

  1. ^ "Our Campaigns - IA US Senate Race - Nov 05, 1968".
  2. ^ Clerk of the United States House of Representatives (1969). "Statistics of the Congressional Election of November 5, 1968" (PDF). U.S. Government Printing Office.
  3. ^ Leip, David. "1968 Senatorial General Election Results - Iowa". US Election Atlas. David Leip's Election Atlas. Retrieved May 20, 2022.