1988 United States Senate election in Ohio

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1988 United States Senate election in Ohio
Flag of Ohio.svg
  1982 November 8, 1988 1994  
  Howard Metzenbaum, 1991.jpg George Voinovich, 1990.jpg
Nominee Howard Metzenbaum George Voinovich
Party Democratic Republican
Popular vote2,480,0381,872,716
Percentage56.97%42.31%

1988 United States Senate election in Ohio results map by county.svg
Ohio 1988 Senate Election By Township.svg
Metzenbaum:      50–60%     60–70%     70–80%
Voinovich:      50–60%     60–70%

U.S. senator before election

Howard Metzenbaum
Democratic

Elected U.S. Senator

Howard Metzenbaum
Democratic

The 1988 United States Senate election in Ohio was held on November 8, 1988. Incumbent Democratic U.S. Senator Howard Metzenbaum won re-election. [1] Metzenbaum easily won the Democratic nomination with over 80% of the vote, while Cleveland Mayor George Voinovich was uncontested in his primary. This was the last U.S. senator to win in the Democratic party at this seat until 2006. Voinovich would later be elected in the other Senate seat ten years later. As of 2024, this remains the last time that Ohio would support different parties in concurrent presidential and Senate elections.

Contents

Major candidates

Democratic

Democratic primary results
PartyCandidateVotes%
Democratic Howard Metzenbaum 1,070,934 83.57%
Democratic Ralph Applegate210,50816.43%
Total votes1,281,442 100.00%

Republican

Republican primary results
PartyCandidateVotes%
Republican George Voinovich 636,806 100.00%
Total votes636,806 100.00%

Results

1988 United States Senate election in Ohio
PartyCandidateVotes%
Democratic Howard Metzenbaum (incumbent) 2,480,038 56.97%
Republican George Voinovich 1,872,71642.31%
Independent David Marshall1510.00%
Majority607,32214.66%
Turnout 4,352,905100.00%
Democratic hold

See also

References

  1. "Our Campaigns - OH US Senate Race - Nov 08, 1988".