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1988 United States Senate election in Ohio

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FieldValue
election_name1988 United States Senate election in Ohio
countryOhio
typepresidential
ongoingno
previous_election1982 United States Senate election in Ohio
previous_year1982
next_election1994 United States Senate election in Ohio
next_year1994
election_dateNovember 8, 1988
image1Howard Metzenbaum, 1991.jpg
nominee1**Howard Metzenbaum**
party1Democratic Party (United States)
popular_vote1**2,480,038**
percentage1**56.97%**
image2George Voinovich, 1990.jpg
nominee2George Voinovich
party2Republican Party (United States)
popular_vote21,872,716
percentage242.31%
map_image{{Switcher
map_size210px
map_caption**Metzenbaum**:
**Voinovich**:
titleU.S. Senator
before_electionHoward Metzenbaum
before_partyDemocratic Party (United States)
after_electionHoward Metzenbaum
after_partyDemocratic Party (United States)

| [[File:1988 United States Senate election in Ohio results map by county.svg|210px]] | County Results | [[File:Ohio 1988 Senate Election By Township.svg|210px]] | Township Results Voinovich:
The 1988 United States Senate election in Ohio was held on November 8, 1988. Incumbent Democratic U.S. Senator Howard Metzenbaum won re-election. 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 time Democrats won this seat until 2006. Voinovich later served as Ohio's Governor from 1991 to 1998 and held the state's other U.S. Senate seat from 1999 to 2011. , 1988 remains the last time that Ohio voted for different parties in concurrent presidential and Senate elections.

Major candidates

Democratic

  • Howard Metzenbaum, incumbent U.S. Senator

Republican

  • George Voinovich, Mayor of Cleveland and former Lieutenant Governor of Ohio

Results

References

References

  1. "Our Campaigns - OH US Senate Race - Nov 08, 1988".
Info: Wikipedia Source

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1988-ohio-elections 1988-united-states-senate-elections united-states-senate-elections-in-ohio
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